How to Prepare for Quantitative Aptitude for CAT (2014)

Block VI: Counting

...BACK TO SCHOOL

In my experience, students can be divided into two broad categories on the basis of their ability in solving chapters of this block.

Category 1: Students who are comfortable in solving questions of this block, since they understand the underlying concepts well.

Category 2: Students who are not able to tackle questions of this block of chapters since they are not conversant with the counting tools and methods in this block.

If you belong to the second category of students, the main thing you would need to do is to familiarise yourself with the counting methods and techniques of Permutations and Combinations. Once you are through with the same, you would find yourself relatively comfortable at both Permutations & Combinations (P&C) and Probability—the chapters in this block which create the maximum problems for students. Set Theory being a relatively easier chapter, the Back to School section of this block concentrates mostly on P & C and Probability. However, before we start looking at these counting methods, right at the outset, I would want you to remove any negative experiences you might have had while trying to study P& C and Probability. So if you belong to the second category of students, you are advised to read on:

Look at the following table:

Suppose, I were to ask you to count the number of cells in the table above, how would you do it??

5 × 5 = 25!!

I guess, all of you realise the fact that the number of cells in this table is given by the product of the number of rows and columns. However, if you ask a 5 year old child to count the same, he would be counting the number of cells physically. In fact, when you were 5 years old, you would also have required to do a physical count of the number of cells in the table. However, there must have come a point where you must have understood that in all such situations (no. of cells in a table, no. of students in a class, etc.) the total count is got by simply multiplying the number of rows and the number of columns. What I am interested in pointing out to you, is that the discovery of this process for this specific counting situation surely made your counting easier!! What used to take you much longer started taking you shorter times. Not only that, in situations where the count was too large (e.g. 100 rows × 48 columns) an infeasible count became extremely easy. So why am I telling you this??

Simply because just as the rows into columns tool for counting had the effect of making your count easier, so also the tools for counting which P& C describe will also have the effect of making counting easier for you in the specific situations of counting that you will encounter. My experience of training students shows that the only reason students have problems in this block is because it has not been explained properly to them in their +2 classes.

If you try to approach these chapters with the approach that it is not meant to complicate your life but to simplify it, you might end up finding out that there is nothing much to fear in this chapter.

THE TOOL BOX APPROACH TO P&C

While studying P&C, your primary objective should be to familiarise yourself with each and every counting situation. You also need to realise that there are a limited number of counting situations which you need to tackle. You can look at this as the process of the creation of what can be described as a counting tool box.

Once this tool box is created, you would be able to understand the basic situations for counting. Then solving tough questions becomes a matter of simplifying the language of the question into the language of the answer—a matter I would come back to later.

Let us now proceed to list out the specific counting situations which you need to get a hold on. You need to know and understand the following twelve situations of counting and the tools that are used in these situations. Please take note that these tools are explained in detail in various parts of the text on the Permutations and Combinations chapter. You are required to keep these in mind along with the specific situations in which these apply. Look at these as a kind of comprehensive list of situations in which you should know how to count using mathematical tools for your convenience.

The Twelve Counting Situations and Their Tools

Tool No. 1: The ⁿC_r tool—This tool is used for the specific situation of counting the number of ways of selecting r things out of n distinct things.

Example: Selecting a team of 11 cricketers out of a team of 16 distinct players will be ¹⁶C₁₁.

Tool No. 2: The tool for counting the number of selections of r things out of n identical things. (Always 1)

Example: The number of ways of selecting 3 letters out of five A’s. Note that in such cases there will only be one way of selecting them.

Tool No. 3: The 2ⁿ tool—The tool for selecting any number (including 0) of things from n distinct things (ⁿC₀ + ⁿC₁ + ⁿC_{2 …..}+ ⁿC_n = 2ⁿ). Example the number of ways you can or cannot eat sweets at a party if there are ten sweets and you have the option of eating one piece of as many sweets as you like or even of not eating any sweet. Will be given by 2¹⁰.

Corollary: The 2ⁿ – 1 tool—Used when zero selections are not allowed. For instance in the above situation if you are asked to eat at least one sweet and rejecting all sweets is not an option.

Tool No. 4: The n + 1 tool—The tool for selecting any number (including 0) of things from n identical things. For instance, if you have to select any number of letters from A, A, A, A and A then you can do it in six ways.

Tool No. 5: The ^n+r–1C_r–1 tool—The tool for distributing n identical things amongst r people/groups such that any person/group might get any number (including 0). For instance, if you have to distribute 7 identical gifts amongst 5 children in a party then it can be done in ^(7+5-1)C_(5-1) ways. i.e. ¹¹C₄ ways.

Tool No. 6: The ^n–1C_r–1 tool—The tool for distributing n identical things amongst r people/groups such that any person/group might get any number (except 0).

Suppose in the above case you have to give at least one gift to each child it can be done in: ⁶C₄ ways.

Tool No. 7: The mnp rule tool—The tool for counting the number of ways of doing three things when each of them has to be done and there are m ways of doing the first thing, n ways of doing the second thing and p ways of doing the last thing. Suppose you have to go from Lucknow to Varanasi to Patna to Ranchi to Jamshedpur and there are 5 trains from Lucknow to Varanasi & there are 4 trains from Varanasi to Patna, six from Patna to Ranchi and 3 from Ranchi to Jamshedpur, then the number of ways of going from Lucknow to Jamshedpur through Varanasi, Patna & Ranchi is 5 × 4 × 6 × 3 = 360.

Note that this tool is extremely crucial and is used to form numbers and words.

For example how many 4 digit numbers can you form using the digits 0, 1, 2, 3, 4, 5 & 6 only without repeating any digits. In this situation the work outline is to choose a digit for the first place (=6, any one out of 1, 2, 3, 4, 5 & 6 as zero cannot be used there), then choosing a digit for the second place (=6, select any one out of 5 remaining digits plus 0), then a digit for the third place (5 ways) and a digit for the fourth place (4 ways).

Tool No 8: The r! tool for arrangement—This tool is used for counting the number of ways in which you can arrange r things in r places. Notice that this can be derived out of the mnp rule tool.

Tool No 9: The AND rule tool—Whenever you describe the counting situation and connect two different parts of the count by using the conjunction ‘AND’, you will always replace the ‘AND’ with a multiplication sign between the two parts of the count. This is also used for solving Probability questions. For instance, suppose you have to choose a vowel ‘AND’ a consonant from the letters of the word PERMIT, you can do it in ⁴C₁ × ²C₁ways.

Tool No 10: The OR rule tool—Whenever you describe the counting situation and connect two different parts of the count by using the conjunction ‘OR, you will always replace the ‘OR’ with an Addition sign between the two parts of the count. This is also used for solving Probability questions. For instance, suppose you have to select either 2 vowels or a consonant from the letters of the word PERMIT, you can do it in ⁴C₂ + ²C₁ ways.

Tool No 11: The Circular Permutation Tool [(n – 1)! Tool]—This gives the number of ways in which n distinct things can be placed around a circle. The need to reduce the value of the factorial by 1 is due to the fact that in a circle there is no defined starting point.

Tool No 12: The Circular Permutation Tool when there is no distinction between clockwise and anti clockwise arrangements [(n – 1)!/2 Tool]—This gives the number of ways in which n distinct things can be placed around a circle when there is no distinction between clockwise and anti clockwise arrangements. In such a case the number of circular permutations are just divided by 2.

Two More Issues

(1) What about the ⁿP_r tool? The ⁿP_r tool is used to arrange r things amongst n. However, my experience shows that the introduction of this tool just adds to the confusion for most students. A little bit of smart thinking gets rid of all the problems that the ⁿP_r tool creates in your mind. For this let us consider the difference between the ⁿP_r and the ⁿC_r tools.

ⁿC_r = n!/[r! × (n – r)!] while ⁿP_r = n!/[(n – r)!].

A closer look would reveal that the relationship between the ⁿP_r and the ⁿC_r tools can be summarised as:

ⁿP_r = ⁿC_r × r!

Read this as: If you have to arrange r things amongst n things, then simply select r things amongst n things and use the r! tool for the arrangement of the selected r items amongst themselves. This will result in the same answer as the ⁿP_r tool.

In fact, for all questions involving arrangements always solve the question in two parts—First finish the selection within the question and then arrange the selected items amongst themselves. This will remove a lot of confusions in questions of this chapter.

(2) Treat Both Permutations and Combinations and Probability as English language chapters rather than Maths chapters: One of the key discoveries in getting stronger at this block is that after knowing the basic twelve counting situations and their tools (as explained above) you need to stop treating the questions in this chapter as questions in Mathematics and rather treat them as questions in English. Thus, while solving a question in these chapters your main concentration and effort should be on converting the question language into the answer language.

What do I mean by this?

Consider this:

Question Language: What is the number of ways of forming 4 letter words from 5 vowels and 6 consonants such that the word contains one vowel and three consonants?

To solve this question, first create the answer language in your mind:

Answer Language: Select one vowel out of 5 AND select three consonants out of 6 AND arrange the four selected letters amongst themselves.

Once you have this language the remaining part of the answer is just a matter of using the correct tools.

Thus, the answer is ⁵C₁ × ⁶C₃ × 4!

As you can see, the main issue in getting the solution to this question was working out the language by connecting the various parts of the solution using the conjunction AND (In this case, the conjunction OR was not required.). Once, you had the language all you had to do was use the correct tools to count.

The most difficult of questions are solved this way. This is why I advise you to treat this chapter as a language chapter and not as a chapter in Mathematics!!

Pre-assessment Test

Directions for Questions 11 and 12: Read the passage below and solve the questions based on it.

India plays two matches each with New Zealand and South Africa. In any match, the probability of different outcomes for India is given below:

Outcome of all the matches are independent of each other.

Chapter 17. PERMUTATIONS AND COMBINATIONS

STANDARD THEORY

Factorial Notation! Or

= n(n – 1) (n – 2) …3.2.1

n! = n(n – 1) (n – 2) …3.2.1

= Product of n consecutive integers starting from 1.

1.0! = 1

2.Factorials of only Natural numbers are defined.

n! is defined only for n ≥ 0

n! is not defined for n < 0

4.ⁿC_r = 1 when n = r.

5.Combinations (represented by ⁿC_r) can be defined as the number of ways in which r things at a time can be SELECTED from amongst n things available for selection.

The key word here is SELECTION. Please understand here that the order in which the r things are selected has no importance in the counting of combinations.

ⁿC_r = Number of combinations (selections) of n things taken r at a time.

ⁿC_r = n! /[r! (n–r)!] ; where n ≥ r (n is greater than or equal to r).

Some typical situations where selection/combination is used:

(a)Selection of people for a team, a party, a job, an office etc. (e.g. Selection of a cricket team of 11 from 16 members)

(b)Selection of a set of objects (like letters, hats, points pants, shirts, etc) from amongst another set available for selection.

In other words any selection in which the order of selection holds no importance is counted by using combinations.

6.Permutations (represented by ⁿP_r) can be defined as the number of ways in which r things at a time can be SELECTED & ARRANGED at a time from amongst n things.

The key word here is ARRANGEMENT. Hence please understand here that the order in which the r things are arranged has critical importance in the counting of permutations.

In other words permutations can also be referred to as an ORDERED SELECTION.

ⁿP_r = number of permutations (arrangements) of n things taken r at a time.

ⁿP_r = n!/ (n – r)!; n ≥ r

Some typical situations where ordered selection/ permutations are used:

(a)Making words and numbers from a set of available letters and digits respectively

(b)Filling posts with people

(c)Selection of batting order of a cricket team of 11 from 16 members

(d)Putting distinct objects/people in distinct places, e.g. making people sit, putting letters in envelopes, finishing order in horse race, etc.)

The exact difference between selection and arrangement can be seen through the illustration below:

Selection

Suppose we have three men A, B and C out of which 2 men have to be selected to two posts.

This can be done in the following ways: AB, AC or BC (These three represent the basic selections of 2 people out of three which are possible. Physically they can be counted as 3 distinct selections. This value can also be got by using 3C₂.

Note here that we are counting AB and BA as one single selection. So also AC and CA and BC and CB are considered to be the same instances of selection since the order of selection is not important.

Arrangement

Suppose we have three men A, B and C out of which 2 men have to be selected to the post of captain and vice captain of a team.

In this case we have to take AB and BA as two different instances since the order of the arrangement makes a difference in who is the captain and who is the vice captain.

Similarly, we have BC and CB and AC and CA as 4 more instances. Thus in all there could be 6 arrangements of 2 things out of three.

This is given by ³P₂ = 6.

7.The Relationship Between Permutation & Combination:

When we look at the formulae for Permutations and Combinations and compare the two we see that,

ⁿP_r = r! × ⁿC_r

= ⁿC_r × ^rP_r

This in words can be said as:

The permutation or arrangement of r things out of n is nothing but the selection of r things out of n followed by the arrangement of the r selected things amongst themselves.

8.MNP Rule: If there are three things to do and there are M ways of doing the first thing, N ways of doing the second thing and P ways of doing the third thing then there will be M × N × P ways of doing all the three things together.The works are mutually inclusive.

This is used to for situations like:

The numbers 1, 2, 3, 4 and 5 are to be used for forming 3 digit numbers without repetition. In how many ways can this be done?

Using the MNP rule you can visualise this as: There are three things to doÆ The first digit can be selected in 5 distinct ways, the second can be selected in 4 ways and the third can be selected in 3 different ways. Hence, the total number of 3 digit numbers that can be formed are 5 × 4 × 3 = 60

9.When the pieces of work are mutually exclusive, there are M+N+P ways of doing the complete work.

Circular Permutations

Consider two situations:

There are three A, B and C. In the first case, they are arranged linearly and in the other, around a circular table –

For the linear arrangement, each arrangement is a totally new way. For circular arrangements, three linear arrangements are represented by one and the same circular arrangement.

So, for six linear arrangements, there correspond only 2 circular arrangements. This happens because there is no concept of a starting point on a circular arrangement. (i.e., the starting point is not defined.)

Generalising the whole process, for n!, there corresponds to be (1/n) n! ways.

Some More Results

1.Number of terms in (a₁ + a₂ + …+a_n)^m is ^{m+n – 1}C_n–1

Illustration: Find the number of terms in (a + b + c)².

Solution: n = 3, m = 2

^{m + n – 1}C_{n – 1} = ⁴C₂ = 6

Corollary: Number of terms in

(1 + x + x² +….xⁿ)^m is mn + 1

2.Number of zeroes ending the number represented by n! = [n /5] + [n /5²] + [n /5³] + …[n /5^x]

[] Shows greatest integer function where 5^x £ n

Illustration: Find the number of zeroes at the end of 1000!

Solution: [1000/5] + [1000/5²] + [1000/5³] + [1000/5⁴]

200 + 40 + 8 + 1 = 249

Corollary: Exponent of 3 in n! = [n!/3] + [n!/32] + [n!/3³] + …[n!/3^x] where 3^x £ n

[] Shows greatest integer fn.

Illustration: Find how many exponents of 3 will be there in 24!.

Solution: [24/3] + [24/3²] = 8 + 2 = 10

3.Number of squares in a square of n × n side = 1² + 2² + 3² + 4² + …. n²

Number of rectangles in a square of n × n side = 1³ + 2³ +3³ + 4³ + …. n³. (This includes the number of squares.)

Thus the number of squares and rectangles in the following figure are given by:

Number of squares = 1² + 2² + 3² = 14 = Œn²

Number of rectangles = 1³ + 2³ + 3³ = 36 (Œn)² = Œn³ for the rectangle.

A rectangle having m rows and n columns:

The number of squares is given by: m.n + (m – 1)(n – 1) + (m – 2)(n – 2) + ….. until any of (m – x) or (n – x) comes to 1.

The number of rectangles is given by: (1 + 2 + …… + m)(1 + 2 + …… + n)

WORKED-OUT PROBLEMS

In the following examples the solution is given upto the point of writing down the formula that will apply for the particular question. The student is expected to calculate the values after understanding the solution.

Problem 17.1 Find the number of permutations of 6 things taken 4 at a time.

Solution The answer will be given by ⁶P₄.

Problem 17.2 How many 3-digit numbers can be formed out of the digits 1, 2, 3, 4 and 5?

Solution Forming numbers requires an ordered selection. Hence, the answer will be ⁵P₃.

Problem 17.3 In how many ways can the 7 letters M, N, O, P, Q, R, S be arranged so that P and Q occupy continuous positions?

Solution For arranging the 7 letters keeping P and Q always together we have to view P and Q as one letter. Let this be denoted by PQ.

Then, we have to arrange the letters M, N, O, PQ, R and S in a linear arrangement. Here, it is like arranging 6 letters in 6 places (since 2 letters are counted as one). This can be done in 6! ways.

However, the solution is not complete at this point of time since in the count of 6! the internal arrangement between P and Q is neglected. This can be done in 2! ways. Hence, the required answer is 6! × 2!.

Task for the student: What would happen if the letters P, Q and R are to be together? (Ans: 5! × 3!)

What if P and Q are never together? (Answer will be given by the formula: Total number of ways – Number of ways they are always together)

Problem 17.4 Of the different words that can be formed from the letters of the words BEGINS how many begin with B and end with S?

Solution B & S are fixed at the start and the end positions. Hence, we have to arrange E, G, I and N amongst themselves. This can be done in 4! ways.

Task for the student: What will be the number of words that can be formed with the letters of the word BEGINS which have B and S at the extreme positions? (Ans: 4! × 2!)

Problem 17.5 In how many ways can the letters of the word VALEDICTORY be arranged, so that all the vowels are adjacent to each other?

Solution There are 4 vowels and 7 consonants in Valedictory. If these vowels have to be kept together, we have to consider AEIO as one letter. Then the problem transforms itself into arranging 8 letters amongst themselves (8! ways). Besides, we have to look at the internal arrangement of the 4 vowels amongst themselves. (4! ways)

Hence Answer = 8! × 4!.

Problem 17.6 If there are two kinds of hats, red and blue and at least 5 of each kind, in how many ways can the hats be put in each of 5 different boxes?

Solution The significance of at least 5 hats of each kind is that while putting a hat in each box, we have the option of putting either a red or a blue hat. (If this was not given, there would have been an uncertainty in the number of possibilities of putting a hat in a box.)

Thus in this question for every task of putting a hat in a box we have the possibility of either putting a red hat or a blue hat. The solution can then be looked at as: there are 5 tasks each of which can be done in 2 ways. Through the MNP rule we have the total number of ways = 2⁵ (Answer).

Problem 17.7 In how many ways can 4 Indians and 4 Nepalese people be seated around a round table so that no two Indians are in adjacent positions?

Solution If we first put 4 Indians around the round table, we can do this in 3! ways.

Once the 4 Indians are placed around the round table, we have to place the four Nepalese around the same round table. Now, since the Indians are already placed we can do this in 4! ways (as the starting point is defined when we put the Indians. Try to visualize this around a circle for placing 2 Indians and 2 Nepalese.)

Hence, Answer = 3! × 4!

Problem 17.8 How many numbers greater than a million can be formed from the digits 1, 2, 3, 0, 4, 2, 3?

Solution In order to form a number greater than a million we should have a 7 digit number. Since we have only seven digits with us we cannot take 0 in the starting position. View this as 7 positions to fill:

_ _ _ _ _ _ _

To solve this question we first assume that the digits are all different. Then the first position can be filled in 6 ways (0 cannot be taken), the second in 6 ways (one of the 6 digits available for the first position was selected. Hence, we have 5 of those 6 digits available. Besides, we also have the zero as an additional digit), the third in 5 ways (6 available for the 2nd position – 1 taken for the second position.) and so on. Mathematically this can be written as:

6 × 6 × 5 × 4 × 3 × 2 × 1 = 6 × 6!

This would have been the answer had all the digits been distinct. But in this particular example we have two 2’s and two 3’s which are identical to each other. This complication is resolved as follows to get the answer:

Problem 17.9 If there are 11 players to be selected from a team of 16, in how many ways can this be done?

Solution ¹⁶C₁₁.

Problem 17.10 In how many ways can 18 identical white and 16 identical black balls be arranged in a row so that no two black balls are together?

Solution When 18 identical white balls are put in a straight line, there will be 19 spaces created. Thus 16 black balls will have 19 places to fill in. This will give an answer of: ¹⁹C₁₆. (Since, the balls are identical the arrangement is not important.)

Problem 17.11 A mother with 7 children takes three at a time to a cinema. She goes with every group of three that she can form. How many times can she go to the cinema with distinct groups of three children?

Solution She will be able to do this as many times as she can form a set of three distinct children from amongst the seven children. This essentially means that the answer is the number of selections of 3 people out of 7 that can be done.

Hence, Answer = ⁷C₃.

Problem 17.12 For the above question, how many times will an individual child go to the cinema with her before a group is repeated?

Solution This can be viewed as: The child for whom we are trying to calculate the number of ways is already selected. Then, we have to select 2 more children from amongst the remaining 6 to complete the group. This can be done in ⁶C₂ ways.

Problem 17.13 How many different sums can be formed with the following coins:

5 rupee, 1 rupee, 50 paisa, 25 paisa, 10 paisa and 1 paisa?

Solution A distinct sum will be formed by selecting either 1 or 2 or 3 or 4 or 5 or all 6 coins.

But from the formula we have the answer to this as : 2⁶ – 1.

[Task for the student: How many different sums can be formed with the following coins:

5 rupee, 1 rupee, 50 paisa, 25 paisa, 10 paisa, 3 paisa, 2 paisa and 1 paisa?

Hint: You will have to subtract some values for double counted sums.]

Problem 17.14 A train is going from Mumbai to Pune and makes 5 stops on the way. 3 persons enter the train during the journey with 3 different tickets. How many different sets of tickets may they have had?

Solution Since the 3 persons are entering during the journey they could have entered at the:

1^st station (from where they could have bought tickets for the 2^nd, 3^rd, 4^th or 5^th stations or for Pune Æ total of 5 tickets.)

2^nd station (from where they could have bought tickets for the 3^rd, 4^th or 5^th stations or for Pune Æ total of 4 tickets.)

3^rd station (from where they could have bought tickets for the 4th or 5th stations or for Pune Æ total of 3 tickets.)

4^th station (from where they could have bought tickets for the 5^th station or for Pune Æ total of 2 tickets.)

5^th station (from where they could have bought a ticket for Pune Æ total of 1 ticket.)

Thus, we can see that there are a total of 5 + 4 + 3 + 2 + 1 = 15 tickets available out of which 3 tickets were selected. This can be done in ¹⁵C₃ ways (Answer).

Problem 17.15 Find the number of diagonals and triangles formed in a decagon.

Solution A decagon has 10 vertices. A line is formed by selecting any two of the ten vertices. This can be done in ¹⁰C₂ ways. However, these ¹⁰C₂ lines also count the sides of the decagon.

Thus, the number of diagonals in a decagon is given by: ¹⁰C₂ – 10 (Answer)

Triangles are formed by selecting any three of the ten vertices of the decagon. This can be done in ¹⁰C₃ ways (Answer).

Problem 17.16 Out of 18 points in a plane, no three are in a straight line except 5 which are collinear. How many straight lines can be formed?

Solution If all 18 points were non-collinear then the answer would have been ¹⁸C₂. However, in this case ¹⁸C₂ has double counting since the 5 collinear points are also amongst the 18. These would have been counted as ⁵C₂whereas they should have been counted as 1. Thus, to remove the double counting and get the correct answer we need to adjust by reducing the count by (⁵C₂ – 1).

Hence, Answer = ¹⁸C₂ – (⁵C₂ – 1) = ¹⁸C₂ – ⁵C₂ + 1

Problem 17.17 For the above situation, how many triangles can be formed?

Solution The triangles will be given by ¹⁸C₃ – ⁵C₃.

Problem 17.18 A question paper had ten questions. Each question could only be answered as True (T) or False (F). Each candidate answered all the questions. Yet, no two candidates wrote the answers in an identical sequence. How many different sequences of answers are possible?

(a)20    (b) 40

(c)512    (d) 1024

Solution 2¹⁰ = 1024 unique sequences are possible. Option (d) is correct.

Problem 17.19 When ten persons shake hands with one another, in how many ways is it possible?

(a)20    (b) 25

(c)40    (d) 45

Solution For n people there are always ⁿC₂ shake hands. Thus, for 10 people shaking hands with each other the number of ways would be ¹⁰C₂ = 45.

Problem 17.20 In how many ways can four children be made to stand in a line such that two of them, A and B are always together?

(a)6    (b) 12

(c)18    (d) 24

Solution If the children are A, B, C, D we have to consider A & B as one child. This, would give us 3! ways of arranging AB, C and D. However, for every arrangement with AB, there would be a parallel arrangement with BA. Thus, the correct answer would be 3! × 2! = 12 ways. Option (b) is correct.

Problem 17.21 Each person’s performance compared with all other persons is to be done to rank them subjectively. How many comparisons are needed to total, if there are 11 persons?

(a)66    (b) 55

(c)54    (d) 45

Solution There would be ¹¹C₂ combinations of 2 people taken 2 at a time for comparison. ¹¹C₂ = 55.

Problem 17.22 A person X has four notes of Rupee 1, 2, 5 and 10 denomination. The number of different sums of money she can form from them is

(a)16    (b) 15

(c)12    (d) 8

Solution 2⁴ – 1 = 15 sums of money can be formed. Option (b) is correct.

Problem 17.23 A person has 4 coins each of different denomination. What is the number of different sums of money the person can form (using one or more coins at a time)?

(a)16    (b) 15

(c)12    (d) 11

Solution 2⁴ – 1 = 15. Hence, option (b) is correct.

Problem 17.24 How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9, such that the digits are in ascending order?

(a)80    (b) 81

(c)83    (d) 84

Solution Numbers starting with 12 – 7 numbers

Numbers starting with 13 – 6 numbers; 14 – 5, 15 – 4, 16 – 3, 17 – 2, 18 – 1. Thus total number of numbers starting from 1 is given by the sum of 1 to 7 = 28.

Number of numbers starting from 2- would be given by the sum of 1 to 6 = 21

Number of numbers starting from 3- sum of 1 to 5 = 15

Number of numbers starting from 4 – sum of 1 to 4 = 10

Number of numbers starting from 5 – sum of 1 to 3 = 6

Number of numbers starting from 6 = 1 + 2 = 3

Number of numbers starting from 7 = 1

Thus a total of: 28 + 21 + 15 + 10 + 6 + 3 + 1 = 84 such numbers. Option (d) is correct.

Problem 17.25 In a carrom board game competition, m boys n girls (m > n > 1) of a school participate in which every student has to play exactly one game with every other student. Out of the total games played, it was found that in 221 games one player was a boy and the other player was a girl.

Consider the following statements:

I.The total number of students that participated in the competition is 30.

II.The number of games in which both players were girls is 78.

Which of the statements given above is/are correct?

(a)I only    (b) II only

(c)Both I and II   (d) Neither I nor II.

Solution The given condition can get achieved if we were to use 17 boys and 13 girls. In such a case both statement I and II are correct. Hence, option (c) is correct.

Problem 17.26

In how many different ways can all of 5 identical balls be placed in the cells shown above such that each row contains at least 1 ball?

(a)64    (b) 81

(c)84    (d) 108

Solution The placement of balls can be 3, 1, 1 and 2, 2, 1. For 3, 1, 1- If we place 3 balls in the top row, there would be ³C₁ ways of choosing a place for the ball in the second row and ³C₁ ways of choosing a place for the ball in the third row. Thus, ³C₁ × ³C₁ = 9 ways. Similarly there would be 9 ways each if we were to place 3 balls in the second row and 3 balls in the third row. Thus, with the 3, 1, 1 distribution of 5 balls we would get 9 + 9 + 9 = 27 ways of placing the balls.

We now need to look at the 2, 2, 1 arrangement of balls. If we place 1 ball in the first row, we would need to place 2 balls each in the second and the third rows. In such a case, the number of ways of arranging the balls would be ³C₁× ³C₂ × ³C₂ = 27 ways. (choosing 1 place out of 3 in the first row, 2 places out of 3 in the second row and 2 places out of 3 in the third row).

Similarly if we were to place 1 ball in the second row and 2 balls each in the first and third rows we would get 27 ways of placing the balls and another 27 ways of placing the balls if we place 1 ball in the third row and 2 balls each in the other two rows.

Thus with a 2, 2, 1 distribution of the 5 balls we would get 27 + 27 + 27 = 81 ways of placing the balls.

Hence, total number of ways = Number of ways of placing the balls with a 3,1,1 distribution of balls + number of ways of placing the balls with a 2, 2, 1 distribution of balls = 27 + 81 = 108.

Hence, option (d) is correct.

Problem 17.27 There are 6 different letter and 6 correspondingly addressed envelopes. If the letters are randomly put in the envelopes, what is the probability that exactly 5 letters go into the correctly addressed envelopes?

(a)Zero    (b) 1/6

(c)1/2    (d) 5/6

Solution If 5 letters go into the correct envelopes the sixth would automatically go into it’s correct envelope. Thus, there is no possibility when exactly 5 letters are correct and 1 is wrong. Hence, option (a) is correct.

Problem 17.28

There are two identical red, two identical black and two identical white balls. In how many different ways can the balls be placed in the cells (each cell to contain one ball) shown above such that balls of the same colour do not occupy any two consecutive cells?

(a)15    (b) 18

(c)24    (d) 30

Solution In the first cell, we have 3 options of placing a ball. Suppose we were to place a red ball in the first cell- then the second cell can only be filled with either black or white – so 2 ways. Subsequently there would be 2 ways each of filling each of the cells (because we cannot put the colour we have already used in the previous cell).

Thus, the required number of ways would be 3 × 2 × 2 × 2 = 24 ways.

Hence, option (c) is correct.

Problem 17.29

How many different triangles are there in the figure shown above?

(a)28    (b) 24

(c)20    (d) 16

Solution Look for the smallest triangles first—there are 12 of them.

Then, look for the triangles which are equal to half the rectangle—there are 12 of them.

Besides, there are 4 bigger triangles (spanning across 2 rectangles).

Thus a total of 28 triangles can be seen in the figure.

Hence, option (a) is correct.

Problem 17.30 A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included a particular student?

(a)6    (b) 8

(c)10    (d) 12

Solution If the students are A, B, C, D, E and F- we can have ⁶C₃ groups in all. However, if we have to count groups in which a particular student (say A) is always selected- we would get ⁵C₂ = 10 ways of doing it. Hence, option (c) is correct.

Problem 17.31 Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2?

(a)36    (b) 81

(c)91    (d) 116

Solution All 3 dice have twos – 1 case.

Two dice have twos:

This can principally occur in 3 ways which can be broken into:

If the first two dice have 2- the third dice can have 1, 3, 4, 5 or 6 = 5 ways.

Similarly, if the first and third dice have 2, the second dice can have 5 outcomes Æ 5 ways and if the second and third dice have a 2, there would be another 5 ways. Thus a total of 15 outcomes if 2 dice have a 2.

With only 1 dice having a two- If the first dice has 2, the other two can have 5 × 5 = 25 outcomes.

Similarly 25 outcomes if the second dice has 2 and 25 outcomes if the third dice has 2. A total of 75 outcomes. Thus, a total of 1 + 15 + 75 = 91 possible outcomes with at least.

Hence, option (c) is correct.

Problem 17.32 All the six letters of the name SACHIN are arranged to form different words without repeating any letter in any one word. The words so formed are then arranged as in a dictionary. What will be the position of the word SACHIN in that sequence?

(a)436    (b) 590

(c)601    (d) 751

Solution All words staring with A, C, H, I and N would be before words starting with S. So we would have 5! Words (= 120 words) each starting with A, C, H, I and N. Thus, a total of 600 words would get completed before we start off with S. SACHIN would be the first word starting with S, because A, C, H, I, N in that order is the correct alphabetical sequence. Hence, Sachin would be the 601^st word. Hence, option (c) is correct.

Problem 17.33 Five balls of different colours are to be placed in three different boxes such that any box contains at least one 1 ball. What is the maximum number of different ways in which this can be done?

(a)90    (b) 120

(c)150    (d) 180

Solution The arrangements can be [3 & 1 & 1 or 1 & 3 & 1 or 1 & 1 & 3] or 2 & 2 & 1 or 2 & 1 & 2 or 1 & 2 and 2.

Total number of ways = 3 × ⁵C₃ × ²C₁ × ¹C₁ + 3 × ⁵C₂ × ³C₂ × ¹C₁ = 60 + 90 = 150 ways

Hence, option (c) is correct.

Problem 17.34 Amit has five friends: 3 girls and 2 boys. Amit’s wife also has 5 friends : 3 boys and 2 girls. In how many maximum number of different ways can they invite 2 boys and 2 girls such that two of them are Amit’s friends and two are his wife’s?

(a)24    (b) 38

(c)46    (d) 58

Solution The selection can be done in the following ways:

2 boys from Amit’s friends and 2 girls from his wife’s friends OR 1 boy & 1 girl from Amit’s friends and 1 boy and 1 girl from his wife’s friends OR 2 girls from Amit’s friends and 2 boys from his wife’s friends.

The number of ways would be:

²C₂ × ²C₂ + ³C₁ × ²C₁ × ³C₁ × ²C₁ + ³C₂ × ³C₂ = 1 + 36 + 9 = 46 ways.

Problem 17.35

In the given figure, what is the maximum number of different ways in which 8 identical balls can be placed in the small triangles 1, 2, 3 and 4 such that each triangle contains at least one ball?

(a)32    (b) 35

(c)44    (d) 56

Solution The ways of placing the balls would be 5, 1, 1, 1 (4!/3! = 4 ways); 4, 2, 1 & 1 (4!/2! = 12 ways); 3, 3, 1, 1 (4!/2! × 2! = 6 ways); 3, 2, 2, 1 (4!/2! = 12 ways) and 2, 2, 2, 2 (1 way). Total number of ways = 4 + 12 + 6 + 12 + 1 = 35 ways. Hence, option (b) is correct.

Problem 17.36 6 equidistant vertical lines are drawn on a board. 6 equidistant horizontal lines are also drawn on the board cutting the 6 vertical lines, and the distance between any two consecutive horizontal lines is equal to that between any two consecutive vertical lines. What is the maximum number of squares thus formed?

(a)37    (b) 55

(c)91    (d) 225

Solution The number of squares would be 1² + 2² + 3² + 4² + 5² + 6² = 91. Hence, option (c) is correct.

Problem 1.37 Groups each containing 3 boys are to be formed out of 5 boys—A, B, C, D and E such that no group contains both C and D together. What is the maximum number of such different groups?

(a)5    (b) 6

(c)7    (d) 8

Solution All groups – groups with C and D together = ⁵C₃ – ³C₁ = 10 – 3 = 7

Problem 17.38

In how many maximum different ways can 3 identical balls be placed in the 12 squares (each ball to be placed in the exact centre of the squares and only one ball is to be placed in one square) shown in the figure given above such that they do not lie along the same straight line ?

(a)144    (b) 200

(c)204    (d) 216

Solution The thought process for this question would be:

All arrangements (¹²C₃) – Arrangements where all 3 balls are in the same row (3 × ⁴C₃)– arrangements where all 3 balls are in the same straight line diagonally (4 arrangements) = ¹²C₃ – 3 × ⁴C₃ – 4 = 220 – 12 – 4 = 204 ways.

Hence, option (c) is correct.

Problem 17.39 How many numbers are there in all from 6000 to 6999 (Both 6000 and 6999 included) having at least one of their digits repeated?

(a)216    (b) 356

(c)496    (d) 504

Solution All numbers – numbers having no numbers repeated = 1000 – 9 × 8 × 7 = 1000 – 504 = 496 numbers. Hence, option (c) is correct.

Problem 17.40 Each of two women and three men is to occupy one chair out of eight chairs, each of which is numbered from one to eight. First, women are to occupy any two chairs from\those numbered one to four; and then the three men would occupy any three chairs out of the remaining six chairs. What is the maximum number of different ways in which this can be done?

(a)40    (b) 132

(c)1440    (d) 3660

Solution ⁴C₂ × 2! × ⁶C₃ × 3! = 6 × 2 × 20 × 6 = 1440. Hence, option (c) is correct.

Problem 17.41 A box contains five set of balls while there are three balls in each set. Each set of balls has one colour which is different from every other set. What is the least number of balls that must be removed from the box in order to claim with certainty that a pair of balls of the same colour has been removed?

(a)6    (b) 7

(c)9    (d) 11

Solution Let C1, C2, C3, C4 and C5 be the 5 distinct colours which have no repetition. For being definitely sure that we have picked up 2 balls of the same colour we need to consider the worst case situation.

Consider the following scenario:

In the above distribution of balls each set has exactly 1 ball which is unique in it’s colour while the colours of the other two balls are shared at least once in one of the other sets. In such a case, the worst scenario would be if we pick up the first 10 balls and they all turn out to be of different colours. The 11^th ball has to be of a colour which has already been taken. Thus,if we were to pick out 11 balls we would be sure of having at least 2 balls of the same colour. Hence, option (d) is correct.

Problem 17.42 In a question paper, there are four multiple-choice questions. Each question has five choices with only one choice as the correct answer. What is the total number of ways in which a candidate will not get all the four answers correct?

(a) 19    (b) 120

(c) 624    (d) 1024

Solution 5⁴ would be the total number of ways in which the questions can be answered. Out of these there would be only 1 way of getting all 4 correct. Thus, there would be 624 ways of not getting all answers correct.

Problem 17.43

Each of 8 identical balls is to be placed in the squares shown in the figure given in a horizontal direction such that one horizontal row contains 6 balls and the other horizontal row contains 2 balls. In how many maximum different ways can this be done?

(a)38    (b) 28

(c)16    (d) 14

Solution The 6 balls must be on either of the middle rows. This can be done in 2 ways. Once, we put the 6 balls in their single horizontal row- it becomes evident that for placing the 2 remaining balls on a straight line there are 2 principal options:

1.Placing the two balls in one of the four rows with two squares. In this case the number of ways of placing the balls in any particular row would be 1 way (since once you were to choose one of the 4 rows, the balls would automatically get placed as there are only two squares in each row.) Thus the total number of ways would be 2 × 4 × 1 = 8 ways.

2.Placing the two balls in the other row with six squares. In this case the number of ways of placing the 2 balls in that row would be ⁶C₂. This would give us ²C₁ × 1 × ⁶C₂ = 30 ways. Total is 30 + 8 = 38 ways.

Hence, option (a) is correct.

Problem 17.44 In a tournament each of the participants was to play one match against each of the other participants. 3 players fell ill after each of them had played three matches and had to leave the tournament. What was the total number of participants at the beginning, if the total number of matches played was 75?

(a)8    (b) 10

(c)12    (d) 15

Solution The number of players at the start of the tournament cannot be 8, 10 or 12 because in each of these cases the total number of matches would be less than 75 (as ⁸C₂, ¹⁰C₂ and ¹²C₂ are all less than 75.) This only leaves 15 participants in the tournament as the only possibility.

Hence, option (d) is correct.

Problem 17.45 There are three parallel straight lines. Two points A and B are marked on the first line, points C and D are marked on the second line and points E and F are marked on the third line. Each of these six points can move to any position on its respective straight line.

Consider the following statements:

I.The maximum number of triangles that can be drawn by joining these points is 18.

II.The minimum number of triangles that can be drawn by joining these points is zero.

Which of the statements given above is/are correct?

(a)I only    (b) II only

(c)Both I and II   (d) Neither I nor II

Solution The maximum triangles would be in case all these 6 points are non-collinear. In such a case the number of triangles is ⁶C₃ = 20. Statement I is incorrect.

Statement II is correct because if we take the position that A and B coincide on the first line, C & D coincide on the second line, E & F coincide on the third line and all these coincidences happen at 3 points which are on the same straight line- in such a case there would be 0 triangles formed. Hence, option (b) is correct.

Problem 17.46 A mixed doubles tennis game is to be played between two teams (each team consists of one male and one female). There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

(a)12    (b) 21

(c) 36    (d) 42

Solution First select the two men. This can be done in ⁴C₂ ways. Let us say the men are A, B, C and D and their respective wives are a, b, c and d.

If we select A and B as the two men then while selecting the women there would be two cases as seen below:

Case 1:

Thus, total number of ways in this case = ⁴C₂ × 1 × ³C₁ = 18 ways.

Case 2:

Total number of ways of doing this = ⁴C₂ × 2 × ²C₁ = 24 ways.

Hence, the required answer is 18 + 24 = 42 ways.

Hence, option (d) is correct.

LEVEL OF DIFFICULTY (I)

Directions for Question 37 to 39: There are 25 points on a plane of which 7 are collinear. Now solve the following:

Directions for Questions 80 and 81: Answer these questions based on the information given below.

Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appear same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters.

Directions for Questions 87 and 88: In a chess tournament there were two women participating and every participant played two games with the other participants. The number of games that the men played among themselves exceeded the number of games that the men played with the women by 66.

LEVEL OF DIFFICULTY (II)

Directions for Questions 41 and 42. There are 40 doctors in the surgical department of the AIIMS. In how many ways can they be arranged to form a team with:

LEVEL OF DIFFICULTY (III)

Directions for Questions 17 and 18: Read the passage below and answer the questions.

In the famous program Kaun Banega Crorepati, the host shakes hand with each participant once, while he shakes hands with each qualifier (amongst participant) twice more. Besides, the participants are required to shake hands once with each other, while the winner and the host each shake hands with all the guests once.

Directions for Questions 37 and 38: In a chess tournament there were two women participating and every participant played two games with the other participants. The number of games that the men played among themselves exceeded the number of games that the men played with the women by 66.

Level of Difficulty (I)

Hints

Level of Difficulty (III)

1.¹⁰C₄ × 6! = 10!/4!

2.For drawing a circle we need 3 non collinear points. This can be done in:

³C₃ + ³C₂ × ⁷C₁ + ³C₁ × ⁷C₂ = 1 + 21 + 63 = 85

3.The odd digits have to occupy even positions. This can be done in = 6 ways.

The other digits have to occupy the other positions. This can be done in = 10 ways.

Hence total number of rearrangements possible = 6 × 10 = 60.

4.The number of straight lines is ⁿC₂ out of which there are n sides. Hence, the number of diagonals is ⁿC₂ – n.

5. ⁿC₂– n= 54.

6.We cannot take ‘0’ since the smallest digit must be placed at the left most place. We have only 9 digits from which to select the numbers. First select any number of digits. Then for any selection there is only one possible arrangement where the required condition is met. This can be done in ⁹C₁ + ⁹C₂ + ⁹C₃ + ... + ⁹C₉ ways = 2⁹ – 1 = 511 ways.

But we can’t take numbers which have only one digit, hence the required answer is 511 – 9.

7.200 runs can be scored by scoring only fours or through a combination of fours and sixes.

Possibilities are 50 × 4, 47 × 4 + 2 × 6, 44 × 4 + 4 × 6 … A total of 17 ways.

8.Of the total arrangements possible (6!) exactly half would have 1 before 6. Thus, 6!/2 = 360.

9.Total number of permutations without any restrictions – Number of permutations having the ‘acd’ pattern – Number of permutations having the ‘beg’ pattern + Number of permutations having both the ‘beg’ and ‘acd’ patterns.

10.A and B can occupy the first and the ninth places, the second and the tenth places, the third and the eleventh place and so on… This can be done in 18 ways.

A and B can be arranged in 2 ways.

All the other 24 alphabets can be arranged in 24! ways.

Hence the required answer = 2 × 18 × 24!

11.First arrange the two sisters around a circle in such a way that there will be one seat vacant between them. [This can be done in 2! ways since the arrangement of the sisters is not circular.]

Then, the other 18 people can be arranged on 18 seats in 18! ways.

12.¹⁰C₂ × ⁸C₁ + ¹⁰C₁ × ⁸C₂ = 360 + 280 = 640

14.A chess board consists 9 parallel lines × 9 parallel lines. For a rectangle we need to select 2 parallel lines and two other parallel lines that are perpendicular to the first set. Hence, ⁹C₂ × ⁹C₂

15-16.Based on direct formulae.

15.This is a direct result based question. Option (c) is correct. Refer to result no. 6 in Important Results 2.

16.(1 + 2 + 3 + … + 18) (1 + 2 + 3 + … + 16)

17-19.Based on simple counting according to the conditions given in the passage

17.¹⁰C₁ + 3 × 2 + ¹⁰C₂ + 60 + 60 = 181

18.Handshakes involving host = 76

Hence, the required ratio is 76: 105.

19.The guests (Spectators) would shake hands ⁶⁰C₂ times = 1770.

Required percentage increase = 977.9%.

20.First select six people out of 12 for the first row. The other six automatically get selected for the second row. Arrange the two rows of people amongst themselves. Besides, the papers can be given in the pattern of 121212 or 212121. Hence the answer is 2 × (¹²C₆ × 6! × 6!).

21.The difference in this question from the previous question is the number of ways in which the papers can be distributed. This can be done by either distributing three different variants in the first three places of each row or by repeating the same variant in the first and the third places.

22.Required permutations = Total permutations with no condition – permutations with the conditions which we do not have to count.

23.We have to count natural numbers which have a maximum of 4 digits. The required answer will be given by:

Number of single digit numbers + Number of two digit numbers + Number of three digit numbers + Number of four digit numbers.

24.Let the three people be A, B and C.

If 1 person gets no objects, the 7 objects must be distributed such that each of the other two get 1 object at least.

This can be done as 6 & 1, 5 & 2, 4 & 3 and their rearrangements.

The answer would be

(⁷C₆ + ⁷C₅ + ⁷C₄) × 3! = 378

Also, two people getting no objects can be done in 3 ways.

Thus, the answer is 378 + 3 = 381

25.If only one gets 1 object

The remaining can be distributed as: (6,0) , (4, 2), (3, 3).

(⁷C₁ × ⁶C₆ × 3! + ⁷C₁ × ⁶C₄ × 3! + ⁷C₁ × ⁶C₃ × 3!/2!)

= 42 + 630 + 420 + 1092.

If 2 people get 1 object each:

⁷C₁ × ⁶C₁ × ⁵C₅ × 3!/2! = 126.

Thus, a total of 1218.

26.Natural numbers which consist of the digits 1, 2, 3, 4. and 5 and are divisible by 4 must have either 12, 24, 32 or 52 in the last two places. For the other two places we have to arrange three digits in two places.

27.No. of 1 digit nos = 9

No. of 2 digit nos = 81

No. of 3 digit nos = 9 × 9 × 8 = 648

No. of 4 digit nos = 9 × 9 × 8 × 7 = 4536

Total nos = 9 + 81 + 648 + 4536 = 5274

28.If the two digits are a and b then 4 digit numbers can be formed in the following patterns.

aabb; aaab or aaaa.

You will have to take two situations in each of the cases- first when the two digits are non zero digits and second when the two digits are zero.

29.For the total of the digits to be odd one of the following has to be true:

The number should contain 1 odd + 6 even or 3 odd + 4 even or 5 odd + 2 even or 7 odd digits. Count each case separately.

33.¹⁸C₄ + ¹⁸C₅ + ¹⁸C₆ + … + ¹⁸C₁₇ + ¹⁸C₁₈

= [¹⁸C₀ + ¹⁸C₄ + … + ¹⁸C₁₈] – [¹⁸C₀ + ¹⁸C₁ + ¹⁸C₂ + ¹⁸C₃]

= 2¹⁸ – [1+ 18 + 153 + 816]

= 261158

35.Total number of 6 digit numbers having 3 odd and 3 even digits (including zero in the left most place) = 5³ × 5³.

From this subtract the number of 5 digit numbers with 2 even digits and 3 odd digits (to take care of the extra counting due to zero)

36.There will be 5 types of numbers, viz. numbers which have

All eight digits even or six even and two odd digits or four even and four odd digits or two even and six odd digits or all eight odd digits. This will be further solved as below:

Eight even digits Æ 5⁸ – 5⁷ = 4 × 5⁷

Six even and two odd digits Æ

when the left most digit is even Æ 4 × ⁷C₅ × 5⁵ × 5⁵

when the left most digit is odd Æ 5 × ⁷C₆ × 5⁶ × 5¹

Four even and four odd digits Æ

when the left most digit is even Æ 4 × ⁷C₅ × 5⁵ × 5⁴

when the left most digit is odd Æ 5 × ⁷C₄ × 5⁴ × 5³

Two even and six odd digits Æ

when the left most digit is even Æ 4 × ⁷C₁ × 5 × 5⁶

when the left most digit is odd Æ 5 × ⁷C₂ × 5² × 5⁵

Eight odd digits Æ 58

37–38.Solve through options.

39.This question is based on a formula: The condition is that ‘n’ things (each thing belonging to a particular place) have to be distributed in ‘n’ places such that no particular thing is arranged in its correct place.

n! – sign of the terms will be alternate and the last term will be .

However, this can also be solved through logic.

40.The possible cases for counting are:

Number of numbers when the units digit is nine + the number of numbers when neither the units digit nor the left most is nine + number of numbers when the left most digit is nine.

42.The condition is that we have to count the number of natural numbers not more than 4300.

The total possible numbers with the given digits = 5 × 5 × 5 × 5 = 625 – 1 = 624.

Subtract form this the number of natural number greater than 4300 which can be formed from the given digits = 1 × 2 × 5 × 5 – 1 = 49.

Hence, the required number of numbers = 624 – 49 = 575.

43.The required answer will be given by

The number of one digit natural number + number of two digit natural numbers + the number of three digit natural numbers + the number of four digit natural number starting with 1, 2, or 3 + the number of four digit natural numbers starting with 4.

46.The following words will appear before SPATE. All words starting with A + All words starting with E + All words starting with P + All words starting with SA + All words starting with SE + SPAET

47.For the maximum possibility assume that no three points other than given in the question are in a straight line.

Hence, the total number of D’s = ^{A + B + C}C₃ – ^AC₃ – ^BC₃ – ^CC₃

48.Use the formula .

n (E) = 7 × 6 × 5 × 4 × 3 × 2 × ⁷C₂

n (S) = 7⁷

n (E) = 8 × 7 + 8 × 7 = 112

n (S) = 64C₂ = =

49.Use the formula

Solutions and Shortcuts

Level of Difficulty (I)

1.The number of numbers formed would be given by 5 × 4 × 3 (given that the first digit can be filled in 5 ways, the second in 4 ways and the third in 3 ways – MNP rule).

2.The first digit can only be 2 (1 way), the second digit can be filled in 7 ways, the third in 6 ways and the fourth in 5 ways. A total of 1 × 7 × 6 × 5 = 210 ways.

3.Each invitation card can be sent in 4 ways. Thus, 4 × 4 × 4 × 4 × 4 × 4 = 4⁶.

4.In this case since nothing is mentioned about whether the prizes are identical or distinct we can take the prizes to be distinct (the most logical thought given the situation). Thus, each prize can be given in 8 ways — thus a total of 8⁵ ways.

5.We need to assume that the 7 Indians are 1 person, so also for the 6 Dutch and the 5 Pakistanis. These 3 groups of people can be arranged amongst themselves in 3! ways. Also, within themselves the 7 Indians the 6 Dutch and the 5 Pakistanis can be arranged in 7!, 6! And 5! ways respectively. Thus, the answer is 3! × 7! × 6! × 5!.

6.Use the MNP rule to get the answer as 5 × 4 = 20.

7.An IITian can make it to IIMs in 2 ways, while a CA can make it through in 3 ways. Required ratio is 2:3. Option (b) is correct.

8.For a straight line we just need to select 2 points out of the 8 points available. ⁸C₂ would be the number of ways of doing this.

9.Use the property ⁿC_r = ⁿC_n-r to see that the two values would be equal at n = 11 since ¹¹C₃ = ¹¹C₈.

10.There would be 5! ways of arranging the 5 letters. Thus, 5! = 120 ways.

11.Rearrangements do not count the original arrangements. Thus, 5!/2! – 1 = 59 ways of rearranging the letters of PATNA.

12.We need to count words starting with P. These words would be represented by P _ _ _ _.

The letters ATNA can be arranged in 4!/2! ways in the 4 places. A total of 12 ways.

13.P _ _ _ T. Missing letters have to be filled with A,N,A. 3!/2!.= 3 ways.

14.Trial and error would give us ⁸C₄ as the answer. ⁸C₄= 8 × 7 × 6 × 5/4 × 3 × 2 ×1 = 70.

15.¹⁰P₃ would satisfy the value given as ¹⁰P₃ = 10 × 9 × 8 = 720.

16.3 × 3 × 2 × 1 = 18

17.3 × 4 × 4 × 4 = 192

18.Divide the numbers into three-digit numbers and 4- digit numbers— Number of 3 digit numbers = 2 × 3 × 2 = 12. Number of 4-digit numbers starting with 10 = 2 × 1 = 2. Total = 14 numbers.

19.3-digit numbers = 2 × 4 × 4 –1 = 31 (–1 is because the number 200 cannot be counted) ; 4-digit numbers starting with 10 = 4 × 4 = 16, Number of 4 digit numbers starting with 11 = 4 × 4 = 16. Total numbers = 31 + 16 + 16 = 63.

20.At n = 3, the values convert to ⁷P₂ and ⁵P₃ whose values respectively are 42 & 60 giving us the required ratio.

21.At r = 7, the value becomes (28!/14! × 14!)/(24!/10! × 14!) Æ 225:11.

22.The maximum value of ⁿC_r for a given value of n, happens when r is equal to the half of n. So if he wants to maximise the number of parties given that he has 20 friends, he should invite 10 to each party.

23.This is a typical case for the use of the formula ^n–1C_r–1 with n = 10 and r = 6. So the answer would be given ⁹C₅.

24.For each digit there would be 5 options (viz 1, 3, 5, 7, 9). Hence, the total number of numbers would be 5 × 5 × 5 × 5 = 625.

25.¹¹C₁ × ¹⁰C₁ = 110. Alternately, ¹¹C₂ × 2!

26.5 × 4 = 20.

27.In the letters of the word ALLAHABAD there is only 1 vowel available for selection (A). Note that the fact that A is available 4 times has no impact on this fact. Also, there are 4 consonants available — viz: L, H, B and D. Thus, the number of ways of selecting a vowel and a consonant would be 1 × ⁴C₁ = 4.

28.Choose 1 person for the single room & from the remaining choose 2 people for the double room & from the remaining choose 4 people for the 4 persons room Æ ⁷C₁ × ⁶C₂ × ⁴C₄.

29.From the first suit there would be 13 options of selecting a card. From the second suite there would be 12 options, from the third suite there would be 11 options and from the fourth suite there would be 10 options for selecting a card. Thus, 13 × 12 × 11 × 10.

30.Number of 11 letter words formed from the letters P, E, R, M, U, T, A, T, I, O, N = 11!/2!.

Number of new words formed = total words –1 = 11!/2!–1.

31.¹⁵C₀ + ¹⁵C₁ + ¹⁵C₂ + ¹⁵C₃ + ¹⁵C₄ + ¹⁵C₅ = 1 + 15 + 105 + 455 + 1365 + 3003 = 4944

32.All arrangements – Arrangements with best and worst paper together = 12! – 2! × 11!.

33.The vowels EUAIO need to be considered as 1 letter to solve this. Thus, there would be 4! ways of arranging Q, T and N and the 5 vowels taken together. Also, there would be 5! ways of arranging the vowels amongst themselves. Thus, we have 4! × 5!.

34.³C₁ × ⁴C₁ × ⁶C₁ = 72.

35.4-digit Motor vehicle registration numbers can have 0 in the first digit. Thus, we have 6 × 5 × 4 × 3 = 360 ways.

36.Single digit numbers = 5

Two digit numbers = 5 × 4 = 20

Three digit numbers = 5 × 4 × 3 = 60

Four digit numbers = 5 × 4 × 3 × 2 = 120

Five digit numbers = 5 × 4 × 3 × 2 × 1 = 120

Total = 5 + 20 + 60 + 120 + 120 = 325

37.²⁵C₂ – ⁷C₂ +1 = 280

38.²⁵C₃ – ⁷C₃ = 2265

39.²⁵C₄ – ⁷C₄ – ⁷C₃ × ¹⁸C₁ = 11985

40.¹⁰C⁵ × 5! = 30240

41.⁸C₃ × ⁵C₂ × 5! = 67200

42.The selection of the 11 player team can be done in ¹⁴C₁₀ ways. This results in the team of 11 players being completely chosen. The arrangements of these 11 players can be done in 11!.

Total batting orders = ¹⁴C₁₀ × 11!= 1001 × 11!

(Note: Arrangement is required here because we are talking about forming batting orders).

43.¹²C₄ = 495

44.³⁰C₃

45.²⁹C₂

46.R _ _ _ _ _ _ W. The letters to go into the spaces are P, O, L, I, N, G. Since all these letters are distinct the number of ways of arranging them would be 6!.

47.7!/3! × 4! = 35

48.The number has to start with a 3 and then in the remaining 6 digits it should have two 3’s and four 0’s. This can be done in 6!/2! × 4! = 15 ways.

49.³C₁ × ⁵C₃ × ⁴C₂ × 5! = 21600

50.If the number of teams is n, then ⁿC₂ should be equal to 45. Trial and error gives us the value of n as 10.

51.From 5 bananas we have 6 choices available (0, 1, 2, 3, 4 or 5). Similarly 4 mangoes and 4 almonds can be chosen in 5 ways each.

So total ways = 6 × 5 × 5 = 150 possible selections. But in this 150, there is one selection where no fruit is chosen.

So required no. of ways = 150 – 1 = 149

Hence Option (b) is correct.

52.For each book we have two options, give or not give. Thus, we have a total of 2¹⁴ ways in which the 14 books can be decided upon. Out of this, there would be 1 way in which no book would be given. Thus, the number of ways is 2¹⁴ – 1.

Hence, Option (d) is correct.

53.The number of ways in which at least 1 Archer book is given is (2⁵–1). Similarly, for Sheldon and Grisham we have (2³–1) and (2⁶–1). Thus required answer would be the multiplication of the three. Hence, Option (d) is the correct answer.

54.For each question we have 3 choices of answering the question (2 internal choices + 1 non-attempt). Thus, there are a total of 3¹⁵ ways of answering the question paper. Out of this there is exactly one way in which the student does not answer any question. Thus there are a total of 3¹⁵ – 1 ways in which at least one question is answered.

Hence, Option (d) is correct.

55.The digits are 1, 6, 7, 8, 7, 6, 1. In this seven-digit no. there are four add places and three even places— OEOEOEO. The four odd digits 1, 7, 7, 1 can be arranged in four odd places in [4!/2! × 2] = 6 ways [as 1 and 7 are both occurring twice].

The even digits 6, 8, 6 can be arranged in three even places in 3!/2! = 3 ways.

Total no. of ways = 6 × 3 = 18.

Hence, Option (c) is correct.

56.We have no girls together, let us first arrange the 5 boys and after that we can arrange the girls in the spaces between the boys.

Number of ways of arranging the boys around a circle = [5 – 1]! = 24.

Number of ways of arranging the girls would be by placing them in the 5 spaces that are formed between the boys. This can be done in ⁵P₃ ways = 60 ways.

Total arrangements = 24 × 60 = 1440.

Hence, Option (d) is correct.

57.Books of interest = 7, books to be borrowed = 3

Case 1— Quants book is taken. Then D.I book can also be taken.

So Amita is to take 2 more books out of 6 which she can do in ⁶C₂ = 15 ways.

Case 2— If Quants book has not been taken, the D.I book would also not be taken.

So Amita will take three books out of 5 books. This can be done in ⁵C₃ = 10 ways.

So total ways = 15 + 10 = 25 ways.

Hence Option (c) is correct.

58.We have to select 5 out of 12.

If Radha and Mohan join- then we have to select only 5 – 2 = 3 dancers out of 12 – 2 = 10 which can be done in ¹⁰C₃ = 120 ways.

If Radha and Mohan do not join, then we have to select 5 out of 12 – 2 = 10-> ¹⁰C₅ = 252 ways.

Total number of ways = 120 + 252 = 372.

Hence, Option (d) is correct.

59. The unit digit can either be 2, 3, 4, 5 or 6.

When the unit digit is 2, the number would be even and hence will be divisible by 2. Hence all numbers with unit digit 2 will be included which is equal to 5! Or 120.

When the unit digit is 3, then in every case the sum of the digits of the number would be 21 which is a multiple of 3. Hence all numbers with unit digit 3 will be divisible by 3 and hence will be included. Total number of such numbers is 5! or 120.

Similarly for unit digit 5 and 6, the number of required numbers is 120 each.

When the unit digit is 4, then the number would be divisible by 4 only if the ten’s digit is 2 or 6. Total number of such numbers is 2 × 4! or 48.

Hence total number of required numbers is (4 × 120) + 48 = 528.

Hence, Option (d) is the answer.

60.As we need to find the maximum number of trials, so we have to assume that the required ball in every box is picked as late as possible. So in the third box, first two balls will be red and black. Hence third trial will give him the required ball. Similarly, in fourth box, he will get the required ball in fourth trial and in the fifth box, he will get the required ball in fifth trial. Hence maximum total number of trials required is 3 + 4 + 5 = 12.

Hence, Option (a) is the answer.

61.Since every player needs to win only 1 match to move to the next round, therefore the 1^st round would have 32 matches between 64 players out of which 32 will be knocked out of the tournament and 32 will be moved to the next round. Similarly in 2^nd round 16 matches will be played, in the 3^rd round 8 matches will be played, in 4^th round 4 matches, in 5^th round 2 matches and the 6^th round will be the final match. Hence total number of rounds will be 6 (2⁶ = 64).

Hence, option (b) is the answer.

62.Total number of pairs of men that can be selected if the adjacent ones are also selected is ^NC₂. Total number of pairs of men selected if only the adjacent ones are selected is N. Hence total number of pairs of men selected if the adjacent ones are not selected is ^NC₂ – N.

Since the total time taken is 88 min, hence the number of pairs is 44.

Hence, ^NC₂ – N = 44 Æ N = 11.

Hence, Option (d) is the answer.

63.Let the number of boys be B. Then ^BC₃ = 36 Æ B = 9

Let the number of girls be G. Then ^GC₂ = 66 Æ G = 12.

Therefore total number of students in the class = 12 + 9 = 21. Hence total number of matches = ²¹C₂ = 210. Hence, number of matches between 1 boy and 1 girl = 210 – (36 + 66) = 108.

Hence, Option (a) is the answer.

64.We will first give 3 chocolates to Sana and 1 each to the other 4. So we have already distributed 7 chocolates. Now we are left with 8 chocolates that have to be distributed among 5 people. So number of ways possible is 12!/8!4! = 495. Out of these we have to eliminate cases in which Sana gets more than 6 chocolates. As we have already given her 3 chocolates that means we have to eliminate cases in which she gets 7 or 8 or 9 or 10 or 11 chocolates out of 495 cases. She would get 11 chocolates in one case, 10 chocolates in ⁴C₁ cases, 9 chocolates in ⁵C₁ cases, 8 chocolates in ⁶C₂ cases and 7 chocolates in ⁷C₃ cases. So, total number of cases that need to be eliminated is 60. So the required number of ways is 495 – 60 = 435.

Hence, Option (d) is the answer.

65.Firstly we will give 5 crores each to the three sons. That will cover 15 crores out of 30 crores leaving behind 15 crores. Now 15 crores can be distributed in three people in 17!15!2! ways or 136 ways.

Hence, Option (a) is the answer.

66.Let x = 3. Then y + z = 27. For the conditions given in the question, no. of solutions is 20.

Similarly for x = 2 there will be 23 solutions, for x = 1 there will be 26 solutions and for x = 0, there will be 29 solutions. Therefore total 98 solutions are possible.

Similarly for y = 3 to 0, there will be 98 solutions and for z = 3 to 0, there will be 98 solutions.

Hence there will be total of 294 solutions. Hence, Option (d) is the answer.

67.No. of words starting with A = 8!/ 2!3! = 3360.

No. of words starting with B = 8!/ 2!4! = 840

No. of words starting with D = 8!/2!4! = 840

No. of words starting with H = 8!/2!4! = 840

Now words with L start.

No. of words starting with LAA = 6!/2! = 180

Now LAB starts and first word starts with LABA.

No. of words starting with LABAA = 4! = 24

After this the next words will be LABADAAHL, LABADAALH, LABADAHAL, LABADAHLA and hence, Option (a) is the answer.

68.We will consider x = 7 to x = 1.

For x = 7, y + z = 5. No. of solutions = 4

For x = 6, y + z = 6. No. of solution = 5

For x = 5, y + z = 7. No. of solutions = 6

For x = 4, y + z = 8. No. of solutions = 7

For x = 3, y = z = 9. No. of solutions = 6

For x = 2, y + z = 10. No. of solutions = 5

For x = 1, y + z = 11. No. of solutions = 4

Hence number of solutions = 37

Hence, Option (b) is the answer.

69.As no three points are collinear, therefore every combination of 3 points out of the nine points will give us a triangle. Hence, the answer is ⁹C₃ or 84.

Hence, Option (d) is correct.

70.The number of combinations of three points picked from the nine given points is ⁹C₃ or 84. All these combinations will result in a triangle except the combination of the three collinear points. Hence number of triangles formed will be 84 – 1 = 83.

Hence, Option (d) is the answer.

71.(xy)² = u! + v

Here xy is a two-digit number and maximum value of its square is 9801. 8! is a five-digit number => u is less than 8 and 4! is 24 which when added to a single digit will never give the square of a two-digit number. Hence u is greater than 4. So, possible values of u can be 5, 6 and 7.

If u = 5, u!= 120 => (xy)² = u! + v => (xy)² = 120 + v = 120 + 1 = 121 = 11²

If u = 6, u! = 720 => (xy)² = u! + v => (xy)² = 720 + v = 720 + 9 = 729 = 27²

If u = 7, u!= 5040 => (xy)² = u! + v => (xy)² = 5040 + v = 5040 + 1 = 5041 = 71²

So there are three cases possible. Hence, 3 solutions exist for the given equation.

Hence, Option (b) is the correct answer.

72.In order to form triangles from the given points, we can either select 2 points from the first line and 1 point from the second OR select one point from the first line and 2 from the second.

This can be done in:

¹⁰C₂ × ¹¹C₁ + ¹⁰C₁ × ¹¹C₂ = 495 + 550 = 1045

73.If we have ‘n’ candidates who can be selected at the maximum, naturally, the answer to the question would also represent ‘n’.

Hence we check for the first option. If n = 3, then 2n + 1 = 7 and it means that there are 7 candidates to be chosen from. Since it is given that the number of ways of selection of at least 1 candidate is 63, we should try to see, whether selecting 1, 2 or 3 candidates from 7 indeed adds up to 63 ways. If it does this would be the correct answer.

⁷C₁ + ⁷C₂ + ⁷C₃ = 7 + 21 + 35 = 63. Thus, the first Option fits the situation and is hence correct.

74.This problem can be approached by putting the white flags in their possible positions. There are essentially 4 possibilities for placing the 3 white flags based on the condition that two flags of the same color cannot be together:

1, 3, 5; 1, 3, 6; 1, 4, 6 and 2, 4, 6.

Out of these 4 possible arrangements for the 3 white flags we cannot use 1, 3, 6 and 1, 4, 6 as these have the same color of flag at both ends- something which is not allowed according to the question. Thus there are only 2 possible ways of placing the white flags— 1, 3, 5 OR 2, 4, 6. In each of these 2 ways, there are a further 3 ways of placing the 1 red flag and the 2 blue flags. Thus we get a total of 6 ways. Option (a) is correct.

75.The possible numbers are: