Top 20 Permutation and Combination Questions With Detailed Video Solutions PDF

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Permutations And Combinations Formula Sheet PDF

The following are some of the important basic formulas of permutations and combinations. You can download these formulas PDF including easy tips and shortcuts to solve the hard questions from this topic. The link to download the PDF is given below.

Permutations and Combinations formula sheet PDF

• Arrangement: n items can be arranged in n! ways, N! = N(N-1)(N-2)(N-3)……1
• 0! = 1! = 1
• Permutation:  A way of selecting and arranging r objects out of a set of n objects, nPr = n!/(n−r)!
• Combination: A way of selecting r objects out of n (arrangement does not matter), nCr = n!/((n−r)!*r!)
• Selecting r objects out of n is the same as selecting (n–r) objects out of n, nCr = nCn–r

Partitioning:

• The number of ways to partition n identical things in r distinct slots is given
  by n+r−1Cr−1
• Number of ways to partition n identical things in r distinct slots so that each slot gets at least 1 is given by n−1Cr−1
• The number of ways to partition n distinct things in r distinct slots is given by rn
• The number of ways to partition n distinct things in r distinct slots where arrangement matters = ((n+r−1)!)/((r−1)!)

Arrangement with repetitions:

• If x items out of n items are repeated, then the number of ways of arranging these n items is n! x! ways. If a items, b items and c items are repeated within n items, they can be arranged in n!/(a! b! c!) ways.

Integral Solutions:

• Number of positive integral solutions to x1+x2+x3+…..+xn= s where s ≥ 0 is s−1Cn−1
• Number of non–negative integral solutions to x1+x2+x3+…..+xn = s where s ≥ 0 is n+s−1Cn−1

Circular arrangement:

• Number of ways of arranging n items around a circle are 1 for n = 1,2 and (n–1)! for n≥3. If its a necklace or bracelet that can be flipped over, the possibilities are ((n−1)!)/2

These are some of the basic formulas for permutations and combinations concepts. One can also check out all the other important formulas and tips to solve the permutations and combinations question in the Cracku’s Permutations and Combinations formula sheet PDF.

Simple Tips and Tricks To Solve CAT Permutation And Combination Questions

• CAT Permutation and Combination  – Tip 1: A few Permutation and Combination CAT concepts appear in the CAT and other MBA entrance exams every year. If you’re starting the prep, firstly understand the CAT Syllabus; Based on our analysis of the Arithmetic CAT Previous Year Questions, this was permutation and combination weightage in CAT: 1 question was asked on this topic (in CAT 2021).
• CAT Permutation and Combination – Tip 2: If you’re not very strong in this topic, learn at least the basics and permutation and combination CAT tricks so that if an easy question comes, you will be able to answer it.  In the CAT exam, CAT P&C questions can be tackled with a strong foundation in this topic. Practice these CAT P&C problems with solutions PDF. Getting yourself acquainted with the basics of these concepts will help you solve the problems. Learn all the major formulae from these concepts. You can check out the permutation and combination CAT study material PDF and learn all the Important CAT Permutation and Combination Formulas pdf here.

Top 20 Permutation and Combination Questions for CAT Preparation

Question 1: In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

a) 200

b) 216

c) 235

d) 256

1) Answer (A)

View Video Solution

Solution:

Number of games in which both the players are girls = $^GC_2$ where G is the number of girls
$^GC_2 = 45$
$^{10}C_2 = 45$
So, G = 10
Similarly, number of games in which both the players are boys = $^BC_2$, where B is the number of boys
$^BC_2 = 190$
$^{20}C_2 = 190$
So, B = 20
So, number of games in which one player is a boy and the other player is a girl is 20*10 = 200

Question 2: If there are 10 positive real numbers $n_1 < n_2 < n_3 … < n_{10}$ , how many triplets of these numbers $(n_1, n_2, n_3 ), ( n_2, n_3, n_4 )$ can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?

a) 45

b) 90

c) 120

d) 180

2) Answer (C)

View Video Solution

Solution:

For any selection of three numbers, there is only one way in which they can be arranged in ascending order.

So, the answer is $^{10}C_3 = 120$

Question 3: A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there has to be exactly 3 girls in the invitees?

a) 320

b) 160

c) 80

d) 200

3) Answer (B)

View Video Solution

Solution:

Selecting 3 girls from 5 girls can be done in $^5C_3$ ways => 10 ways

Each of the boys may or may not be selected => 2 * 2 * 2 * 2 = 16 ways

=> 16 * 10 = 160 ways

Question 4: For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

a) 3

b) 4

c) 2

d) 5

4) Answer (A)

View Video Solution

Solution:

At least one candidate and at most n candidates among 2n+1 candidates =>

^{2n+1}C_1Total number of ways alteast one candidate can be selected = Total number of ways of selecting a candidate – total number of ways of selecting no candidates

$\Rightarrow ^{2n+1}C_1+^{2n+1}C_2+^{2n+2}C_3+….+^{2n+1}C_{n-1}+^{2n+1}C_n = 63 $

We know that $^{2n+1} C_0$ and $^{2n+1} C_{2n+1}$ are equal to 1.

By Binomial Expansion

$ ^{2n+1}C_0+^{2n+1}C_1$+… + $^{2n+1}C_{2n}+^{2n+1}C_{2n+1} = (1+1)^{2n+1} $ —- Eq 1

Also $^{2n+1}C_1$ = $^{2n+1}C_{2n}$ by symmetry

and $^{2n+1}C_2$ = $^{2n+1}C_{2n-1}$ and so on

So $\Rightarrow ^{2n+1}C_{n+1}+^{2n+1}C_{n+2}+….+^{2n+1}C_{2n-1}+^{2n+1}C_{2n} = 63 $

Therefore, on substituting these values in Eq 1 we get

1 + 63 + 63 +1 = $2^{2n+1}$

$2^{2n+1}$ = 128

2n+1 = 7

Therefore, n=3

As at most n students can be selected, the correct answer is 3.

Question 5: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

5) Answer: 160

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Solution:

The total number of given points are 11. (10 on circumference and 1 is the center)
So total possible triangles = 11C3 = 165.
However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 – 5 = 160

Question 6: How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

a) 499

b) 500

c) 375

d) 376

6) Answer (D)

View Video Solution

Solution:

We have to essentially look at numbers between 1000 and 4000 (including both).

The first digit can be either 1 or 2 or 3.

The second digit can be any of the five numbers.

The third digit can be any of the five numbers.

The fourth digit can also be any of the five numbers.

So, total is 3*5*5*5 = 375.

However, we have ignored the number 4000 in this calculation and hence the total is 375+1=376

Question 7: What is the number of distinct terms in the expansion of $(a + b + c)^{20}$?

a) 231

b) 253

c) 242

d) 210

e) 228

7) Answer (A)

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Solution:

The power is 20.

20 has to be divided among a, b and c. This can be done in $^{20+3-1}C_{3-1}$ = $^{22}C_2$ = 231

Option a) is the correct answer.

Question 8: There are 6 boxes numbered 1,2,… 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

a) 5

b) 21

c) 33

d) 60

8) Answer (B)

View Video Solution

Solution:

If there is only 1 green ball, it can be done in 6 ways
If there are 2 green balls, it can be done in 5 ways.
.
.
.
If there are 6 green balls, it can be done in 1 way.
So, the total number of possibilities is 6*7/2 = 21

Question 9: In the adjoining figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

a) 15

b) 35

c) 120

d) 336

9) Answer (B)

View Video Solution

Solution:

The person has to take 3 steps north and 4 steps west, in whatever way he travels.

Total steps = 7, 3 north and 4 west.

Number of ways = 7!/(4!3!) = 35

Question 10: A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is

a) 12 × 81

b) 16 × 192

c) 20 × 125

d) 24 × 216

10) Answer (A)

View Video Solution

Solution:

The number of ways of selecting a colour for the first stripe is 4. The number of ways of selecting a colour for the second stripe is 3. Similarly, the number of ways of selecting colours for the third, fourth, fifth and sixth stripes are 3, 3, 3 and 3 respectively.
The total number of ways of selecting the colours is, therefore, 4*3*3*3*3*3 = 12*81.

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Question 11: Consider the set S = { 1, 2, 3, .., 1000 }. How many arithmetic progressions can be formed from the elements of S that start with l and end with 1000 and have at least 3 elements?

a) 3

b) 4

c) 6

d) 7

e) 8

11) Answer (D)

View Video Solution

Solution:

The nth term is a + (n-1)d
1000 = 1 + (n-1)d
So, (n-1)d = 999
999 = 3^3 * 37
So, the number of factors is 4*2 = 8
Since there should be at least 3 terms in the series, d cannot be 999.
So, the number of possibilities is 7

Question 12: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

a) 16

b) 20

c) 14

d) 15

12) Answer (A)

View Video Solution

Solution:

We have been given that a + b + c + d = 7
Total ways of distributing 7 things among 4 people so that each one gets at least one = $^{n-1}C_{r-1}$ = 6C3 = 20
Now we need to subtract the cases where any one person got more than 3 erasers. Any person cannot get more than 4 erasers since each child has to get at least 1. Any of the 4 childs can get 4 erasers. Thus, there are 4 cases. On subtracting these cases from the total cases we get the required answer. Hence, the required value is 20 – 4 = 16

Question 13: The numbers 1, 2, …, 9 are arranged in a 3 X 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.
If the top left and the top right entries of the grid are 6 and 2, respectively, then the bottom middle entry is

13) Answer: 3

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Solution:

According to the question each column, each row, and each of the two diagonals of the 3X3 matrix add up to the same value. This value must be 15.

Let us consider the matrix as shown below:

Now we’ll try substituting values from 1 to 9 in the exact middle grid shown as ‘x’.

If x = 1 or 3, then the value in the left bottom grid will be more than 9 which is not possible.

x cannot be equal to 2.

If x = 4, value in the left bottom grid will be 9. But then addition of first column will come out to be more than 15. Hence, not possible.

If x=5, we get the grid as shown below:

Hence, for x = 5 all conditions are satisfied. We see that the bottom middle entry is 3.

Hence, 3 is the correct answer.

Question 14: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

14) Answer: 6

View Video Solution

Solution:

After Amal, Bimal and Kamal are given their minimum required pens, the pens left are 8 – (1 + 2 + 3) = 2 pens
Now these two pens have to be divided between three persons so that each person can get zero pens = $^{2+3-1}C_{3-1}$ = $^4C_2$  = 6

Question 15: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

15) Answer: 50

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Solution:

For the number to be divisible by 6, the sum of the digits should be divisible by 3 and the units digit should be even. Hence we have the digits as

Case I: 2, 3, 4, 6
Now the units place can be filled in three ways (2,4,6), and the remaining three places can be filled in 3! = 6 ways.
Hence total number of ways = 3*6 = 18

Case II: 0, 2, 3, 4
case II a: 0 is in the units place => 3! = 6 ways
case II b: 0 is not in the units place => units place can be filled in 2 ways( 2,4) , thousands place can be filled in 2 ways ( remaining 3 – 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 2 * 2 * 2 = 8
Total number of ways in this case = 6 + 8 = 14 ways.

Case III: 0, 2, 4, 6
case III a: 0 is in the units place => 3! = 6 ways

case II b: 0 is not in the units place => units place can be filled in 3 ways( 2,4,6) , thousands place can be filled in 2 ways (remaining 3 – 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 3 * 2 * 2 = 12
Total number of ways in this case = 6 + 12= 18 ways.

Hence the total number of ways = 18 + 14  + 18 = 50 ways

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Question 16: How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

16) Answer: 502

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Solution:

It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number.
There are 9 numbers (1,2,3,4,5,6,7,8,9) in total.
2 digit numbers can be formed in $9C_2$ ways.
3 digit numbers can be formed in $9C_3$ ways.
…………………………………………..
..9 digit number can be formed in 9C9 ways.

We know that $nC_0+nC_1+nC_2+………+nC_n =2^n$
=> $9C_0 + 9C_1+9C_2+….9C_9 = 2^9$
$9C_0 + 9C_1 + …9C_9 = 512$

We have to subtract $9C_0$ and $9C_1$ from both the sides of the equations since we cannot form single digit numbers.
=> $9C_2 + 9C_3+…+9C_9=512-1-9$
$9C_2+9C_3+…+9C_9=502$

Therefore, $502$ is the right answer.

Question 17: How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?

17) Answer: 6

View Video Solution

Solution:

Let ‘ab’ be the two digit number. Where b $\neq$ 0.

We will get number ‘ba’ after interchanging its digit.

It is given that 10a+b > 3*(10b + a)

7a > 29b

If b = 1, then a = {5, 6, 7, 8, 9}

If b = 2, then a = {9}

If b = 3, then no value of ‘a’ is possible. Hence, we can say that there are a total of 6 such numbers.

Question 18: In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is

18) Answer: 1098

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Solution:

In a tournament, there are 43 junior level and 51 senior level participants.

Let ‘n’ be the number of girls on junior level. It is given that the number of girl versus girl matches in junior level is 153.

$\Rightarrow$ nC2 = 153

$\Rightarrow$ n(n-1)/2 = 153

$\Rightarrow$ n(n-1) = 306

=> n$^{2}$-n-306 = 0

=> (n+17)(n-18)=0

=> n=18  (rejecting n=-17)

Therefore, number of boys on junior level = 43 – 18 = 25.

Let ‘m’ be the number of boys on senior level. It is given that the number of boy versus boy matches in senior level is 276.

$\Rightarrow$ mC2 = 276

$\Rightarrow$ m = 24

Therefore, number of girls on senior level = 51 – 24 = 27.

Hence, the number of matches a boy plays against a girl  = 18*25+24*27 = 1098

Question 19: With rectangular axes of coordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y+1) or to (x+1, y), is

19) Answer: 3920

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Solution:

The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)

To calculate the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y direction

Hence the number of paths from (1,1) to (4,6) = $^{(3+5)}C_3$ = 56

To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y direction

Hence the number of paths from (4,6) to (8,10) = $^{(4+4)}C_4$ = 70

The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920

Question 20: The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

20) Answer: 47

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Solution:

The possible arrangements are of the form

35 _ Can be chosen in 6 ways.

35 _ _ We can choose 2 out of the remaining 6 in $^6C_2=15$ways. We remove 1 case where 7 and 8 are together to get 14 ways.

35 _ _ _We can choose 3 out of the remaining 6 in $^6C_3=20$ways. We remove 4 cases where 7 and 8 are together to get 16 ways.

35 _ _ _ _We can choose 4 out of the remaining 6 in $^6C_4=15$ways. We remove 6 case where 7 and 8 are together to get 9 ways.

35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.

Thus, total number of cases = 6+14+16+9+2 = 47.

Alternatively,

The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.

Considering the cases, only 7 is selected.

We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.

Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have

Each of the numbers can either be selected or not selected and we have 4 numbers :

Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.

SImilarly, including only 8, we have 16 more possibilities.

Cases including neither 7 nor 8.

We must have 3 and 5 in the group but there must be no 7 and 8 in the group.

Hence we have 3 5 _ _ _ _.

For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.

= 16 possibilities

But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.

Hence 16 – 1 = 15 cases.

Total = 16+15+16 = 47 possibilities