Chapter 16: PERMUTATION AND COMBINATION

The answer will be given by
6P4

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

n 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!.

5! × 3!

(Answer will be given by the formula: Total number of ways –
Number of ways they are always together)

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.

4! × 2!

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!

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).

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!

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: (6 * 6)/(2! * 2!)

16C11

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.)

 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

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

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.]

Since the 3 persons are entering during the journey they could have entered at the:
1st station (from where they could have bought tickets for the 2nd, 3rd, 4th or 5th stations or for Pune total of 5 tickets.)
2nd station (from where they could have bought tickets for the 3rd, 4th or 5th stations or for Pune Æ total of 4 tickets.)
3rd station (from where they could have bought tickets for the 4th or 5th stations or for Pune Æ total of 3 tickets.)
4th station (from where they could have bought tickets for the 5th station or for Pune Æ total of 2 tickets.)
5th 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).

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).

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 5C2 whereas 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

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

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

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.

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

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

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

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

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

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.

 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.

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.

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.

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

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

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.

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.

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.

n 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.

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.

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

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

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

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

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

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.

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.

 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.

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

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 and12C2 are all less than 75.)
This only leaves 15 participants in the tournament as the only possibility.
Hence, option (d) is correct.

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

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:


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.

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).

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.

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

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.

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!.

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

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. 

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.

Use the property
nCr = nCn-r
to see that the two values would be equal at n = 11 since
11C3 = 11C8

Solve this one through options. If you pick up option (a) it gives you 12 participants in the
tournament. This means that there are 10 men and 2 women. In this case there would be 2 ×
10C2 = 90 matches amongst the men and 2 × 10C1 × 2C1 = 40 matches between 1 man and 1 woman. The
difference between number of matches where both participants are men and the number of matches
where 1 participant is a man and one is a woman is 90 – 40 = 50 – which is not what is given in
the problem.
With 15 participants Æ 11 men and 2 women.
In this case there would be 2 × 11C2 = 110 matches amongst the men and 2 × 11C1 × 2C1 = 44
matches between 1 man and 1 woman. The difference between number of matches where both
participants are men and the number of matches where 1 participant is a man and one is a woman
is 110 – 44 = 66 – which is the required value as given in the problem. Thus, option (b) is correct

Based on the above thinking we get that since there are 15 players and each player plays each of
the others twice, the number of games would be 2 × 15C2 = 2 × 105 = 210

Number of even numbers = 6 × 6 × 6 × 3 = 648

For each selection there are 3 ways of doing it. Thus, there are a total of 3 × 3 × 3 × 3 × 3 = 243.

One digit no. = 5; Two digit nos = 5 × 4 = 20; Three digit no = 5 × 4 × 3 = 60; four digit no = 5 ×
4 × 3 × 2 = 120; Five digit no.= 5 × 4 × 3 × 2 × 1 = 120 Total number of nos = 325

With one green ball there would be six ways of doing this. With 2 green balls 5 ways, with 3
green balls 4 ways, with 4 green balls 3 ways, with 5 green balls 2 ways and with 6 green balls 1
way. So a total of 1 + 2 + 3 + 4 + 5 + 6 = 21 ways

Each of the first, third and fourth options can be obviously seen to be true— no mathematics
needed there. Only the second option can never be true.
In order to think about this mathematically and numerically— think of a party of 3 persons say A, B
and C. In order for the second condition to be possible, each person must know a different number
of persons. In a party with 3 persons this is possible only if the numbers are 0, 1 and 2. If A knows
both B and C (2), B and C both would know at least 1 person— hence it would not be possible to
create the person knowing 0 people. The same can be verified with a group of 4 persons i.e., the
minute you were to make 1 person know 3 persons it would not be possible for anyone in the
group to know 0 persons and hence you would not be able to meet the condition that every person
knows a different number of persons

Since there are 11 symmetric letters, the number of passwords that can be formed would be 11 ×
10 × 9 × 8 = 7920.

This would be given by the number of passwords having:
1 symmetric and 2 asymmetric letters + 2 symmetric and 1 asymmetric letter + 3 symmetric and 0
asymmetric letters
11C1 × 15C2 × 3! + 11C2 × 15C1 × 3! + 11C3 × 3! = 11 × 105 × 6 + 55 × 15 × 6 + 11 × 10 × 9 = 6930
+ 4950 + 990 = 12870

The white square can be selected in 32 ways and once the white square is selected 8 black
squares become ineligible for selection. Hence, the black square can be selected in 24 ways. 32 ×
24 = 768.

A million is 1000000 (i.e. the first seven digit number). So we need to find how many numbers of
less than 7 digits can be formed using the digits 0,7 and 8.
Number of 1 digit numbers = 2
Number of 2 digit numbers = 2 × 3 = 6
Number of 3 digit numbers = 2 × 3 × 3 = 18
Number of 4 digit numbers = 2 × 3 × 3 × 3 = 54
Number of 5 digit numbers = 2 × 3 × 3 × 3 × 3 = 162
Number of 6 digit numbers = 2 × 3 × 3 × 3 × 3 × 3 = 486
Total number of numbers = 728

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.

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

For each book we have two options, give or not give. Thus, we have a total of 2
14 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

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 (25–1)(23–1)(26–1)

For each question we have 3 choices of answering the question (2 internal choices + 1 nonattempt).
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

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

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

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.

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.

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.

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.

Since every player needs to win only 1 match to move to the next round, therefore the 1st
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 2nd round 16 matches will be played, in the
3rd round 8 matches will be played, in 4th round 4 matches, in 5th round 2 matches and the 6th
round will be the final match. Hence total number of rounds will be 6 (26 = 64).

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 so N = 11