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# CAT Permutation and Combination PDF [Most Important]

Permutation and Combination is one of the important topics in the CAT Quantitative Ability (QA) section. If you are not thorough with these topics, make sure you are at least aware of the basics so that you can solve the easy questions in the exam. So ensure that you are aware of the Important ways to solve Permutation and Combination in QA. You can check out these CAT Permutation and Combination questions from the CAT Previous year papers. This post will look into some important Permutations and Combinations for CAT. These are a good source of practice for CAT exam; If you want to practice these questions, you can download these Important CAT Permutation and Combination Questions PDF (with detailed answers) below, which is completely Free.

Question 1: 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

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 2: 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

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 3: 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

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 4: 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

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 5: 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

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.

Question 6: 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

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 7: 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

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 8: 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

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 9: 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

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 10: 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

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.