- Permutation : It is the different arrangements of a given number of elements taken one by one, or some, or all at a time. For example, if we have two elements A and B, then there are two possible arrangements, AB and BA.
- Number of permutations when ‘r’ elements are arranged out of a total of ‘n’ elements is
^{n}**P**. For example, let n = 4 (A, B, C and D) and r = 2 (All permutations of size 2). The answer is 4!/(4-2)! = 12. The twelve permutations are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB and DC._{r}= n! / (n – r)! - Combination : It is the different selections of a given number of elements taken one by one, or some, or all at a time. For example, if we have two elements A and B, then there is only one way select two items, we select both of them.
- Number of combinations when ‘r’ elements are selected out of a total of ‘n’ elements is
^{n}C_{r}= n! / [ (r !) x (n – r)! ]. For example, let n = 4 (A, B, C and D) and r = 2 (All combinations of size 2). The answer is 4!/((4-2)!*2!) = 6. The six combinations are AB, AC, AD, BC, BD, CD. ^{n}C_{r}=^{n}C_{(n – r)}- Permutation and Combination | Set-2
- LCM and HCF
- Work and Wages
- Pipes and Cisterns
- Trains, Boats and Streams
- Ratio Proportion and Partnership
- Mixture and Alligation
- Profit and Loss
- Trigonometry & Height and Distances
- Assumptions and Conclusions, Courses of Action
- Error Detection and Correction
- Trigonometry & Height and Distances | Set-2
- Cvent Interview Experience (On campus for Internship and Full Time)
- TCS Ninja Interview Experience and Interview Questions
- Problem on HCF and LCM
- Problems on Work and Wages
- Problem on Pipes and Cisterns
- Problem on Time Speed and Distance
- Problem on Trains, Boat and streams
- Ratio proportion and partnership | Set-2

NOTE : In the same example, we have different cases for permutation and combination. For permutation, AB and BA are two different things but for selection, AB and BA are same.

### Sample Problems

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

**Solution : **The word “DELHI” has 5 different words.

Therefore, required number of words = ^{5} P _{3} = 5! / (5 – 3)!

=> Required number of words = 5! / 2! = 120 / 2 = 60

**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 type of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character.

So, now we have a total of 5 characters in the word, namely, D, R, V, R, 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

**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 = ^{15} C _{4} = 15 ! / [(4 !) x (11 !)]

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

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

**Solution : **Number of ways 3 boys can be selected out of 6 = ^{6} C _{3} = 6 ! / [(3 !) x (3 !)] = (6 x 5 x 4) / (3 x 2 x 1) = 20

Number of ways 2 girls can be selected out of 5 = ^{5} C _{2} = 5 ! / [(2 !) x (3 !)] = (5 x 4) / (2 x 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 a total of 5 characters in the word, namely, D, R, V, R, 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

Also, total number of possible words = 6! / 2! = 720 / 2 = 360

Therefore, total number of possible words such that the vowels are never together = 360 – 120 = 240

### Problems on Permutation and Combination | Set-2

Quiz on Permutation and Combination

Practice Questions on Permutation and Combination.

This article has been contributed by **Nishant Arora**. Please write comments if you have any doubts related to the topic discussed above, or if you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.

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