**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

**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 & 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