Permutation and Combination - Solved Questions and Answers

Permutation is defined as the Arrangement of items where order matters, whereas Combination is defined as the selection of items where order doesn’t matter. 

Permutations and Combinations questions and answers are provided below for you to learn and practice.

Question 1: How many words can be formed by using 3 letters from the word "DELHI"? 

Solution: 

Here we will use the Permutations for this question. Formula for permutation is nPr, for this we have,

n = 5: Total 5 Letters

r = 3: Letters word we required

ⁿP_r = n!/(n-r)!

⁵P₃ = 5!/2! = 120/2 = 60

So, Total we can form 60 different permutation of word from Letter Delhi.

Question 2: How many words can be formed by using the letters from the word "DRIVER" such that all the vowels are always together? 

Solution: 

In these types of questions, we assume all the vowels to be a single character, i.e., "IE" is a single character.

So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times.

Number of possible arrangements = 5! / 2! = 60

Now, the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together = 60 × 2 = 120 

Question 3: In how many ways, can we select a team of 4 students from a given choice of 15?

Solution: 

Number of possible ways of selection = ¹⁵C₄ = 15 ! / ((4 !) × (11 !))

Number of possible ways of selection = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) = 1365 

Question 4: In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 boys and 2 girls out of 5 girls?

Solution: 

Number of ways 3 boys can be selected out of 6 = ⁶ C ₃ = 6 ! / [(3 !) × (3 !)] = (6 × 5 × 4) / (3 × 2 × 1) = 20

Number of ways 2 girls can be selected out of 5 = ⁵ C ₂ = 5 ! / [(2 !) × (3 !)] = (5 × 4) / (2 × 1) = 10

Therefore, total number of ways of forming the group = 20 x 10 = 200 

Question 5: How many words can be formed by using the letters from the word "DRIVER" such that all the vowels are never together?

Solution: 

We assume all the vowels to be a single character, i.e., "IE" is a single character.

So, now we have 5 characters in the word, namely, D, R, V, R, and IE.

But, R occurs 2 times.

Number of possible arrangements = 5! / 2! = 60

Now, the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together = 60 x 2 = 120 ,

Total number of possible words = 6! / 2! = 720 / 2 = 360

Therefore, the total number of possible words such that the vowels are never together 240

Question 6: How many words can be formed by using 4 letters from the word "COMPUTER"?

Solution:

For this question, we will use Permutations.

n=8: Total 8 letters in the word "COMPUTER"

r=4: We are required to form a 4-letter word

Using the permutation formula:

r = \frac{n!}{(n - r)!}

8P4 = \frac{8!}{(8 - 4)!} = \frac{8!}{4!} = \frac{40320}{24} = 1680

So, the total number of different permutations of words we can form from the letters in "COMPUTER" is 1680.

Question 7: How many words can be formed by using the letters from the word "BALLOON" such that all the vowels are always together?

Solution:

We will assume all vowels (A and O) together as a single unit, i.e., "AO" becomes a single character.

Now, we have the characters B, L, L, O, N, and AO. But, L occurs 2 times.

Number of possible arrangements = \frac{5!}{2!} = \frac{120}{2} = 60

Now, the vowels (A and O) can be arranged in 2!=2 ways.

Therefore, the total number of possible words where the vowels are always together is:

60 \times 2 = 120

So, the total number of possible words is 120.

Question 8: In how many ways can we select a team of 5 students from a given choice of 20?

Solution:

The number of possible ways of selection is given by:

20C5 = \frac{20!}{5!(20 - 5)!} = \frac{20!}{5! \times 15!}

Simplifying:

\frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 15504

So, the number of ways to select 5 students from 20 is 15504.

Tricks to Solve Permutation & Combination Questions