### CLAT Maths: Solved Questions of Permutation and Combination

1. A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters ?

a. 729

b. 625

c. 676

d. 576

Solution: Choose 5 starters from a team of 12 players. Order is not important. C(12,5) = 729 Permutations and Combinations for CLAT

2. How many 4- letter words with or without meaning, can be formed out of the letters of the word, & Logarithms, if repetition of letters is not allowed?

a. 4050

b. 3600

c. 2400

d. 5040

Solution: Logarithms contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= 10P4

= 5040.

3. A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

a. 1260

b. 1400

c. 1600

d. 1320

Solution: A team of 6 members has to be selected from the 10 players. This can be done in C(10,6) or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 210×6= 1260

4. In how many different ways can the letters of the word OPTICAL be arranged so that the vowels always come together?

a. 120

b. 360

c. 720

d. 480

Solution:

The word OPTICAL contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

5. A committee of 5 persons is to be formed from 6 foreigner and 4 natives. In how many ways can this be done when at least 2 natives are included?

a. 196

b. 190

c. 186

d. 200

Solution: There can be three ways- 3 foreigners &amp; 2 natives, 2 foreigners and 3 natives, 1 foreigner and 4 natives

6C3×2C4 + 6C2×4C3 + 6C1×4C4

6 x 20 + 4 x 15 + 1x 6 = 120 + 60 + 6 =186

6. In how many ways can the letters of the word LEADER be arranged?

a. 120

b. 360

c. 480

d. 720

Solution:

The word LEADER has 6 letters.

But in these 6 letters, E occurs 2 times and rest of the letters are different.

Hence, number of ways to arrange these letters =6! / 2! = 6×5×4×3×2/2 = 360

7. How many arrangements of the letters of the word ‘BENGALI’ can be made if the vowels are to occupy only odd places.

a. 720

b. 576

c. 625

d. 567

Solution:

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.

There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4P4 ways and 4 constants can be arranged in 4P4 ways.

Number of words = 4P3 x 4P4 = 24 x 24 = 576

8. A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

a. 126

b. 120

c. 156

d. 170

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows:

Case I: 5 students in the first car and 3 in the second

Case II: 4 students in the first car and 4 in the second

Therefore, the total number of ways in which 8 students can travel is: C(8,3) + C(8,4) = 56 + 70= 126

9. A coin is tossed 3 times. Find out the number of possible outcomes.

a. 9

b. 8

c. 1

d. 3

When a coin is tossed once, there are two possible outcomes: Head(H) and Tale(T)

Hence, when a coin is tossed 3 times, the number of possible outcomes =2×2×2=8=2×2×2=8

10. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw?

a. 64

b. 128

c. 32

d. 64

Solution:

From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that at least one black ball should be there.

Number of ways to select 3 black balls = 3C3

Number of ways to select 2 black balls and 1 non-black ball = 3C2 × 6C1

Number of ways to select 1 black ball and 2 non-black balls = 3C1 × 6C2

Total number of ways = 3C3 + 3C2 × 6C1 + 3C1 × 6C2 =1+3×6+3×6×52×1=1+18+45=64=1+3×6+3×6×52×1=1+18+45=64