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Mathematics | Problems On Permutations | Set 1
• Difficulty Level : Medium
• Last Updated : 16 Oct, 2020

Prerequisite – Permutation and Combination

Formula’s Used :

```1. P(n, r) = n! / (n-r)!

2. P(n, n) = n!  ```

Example-1 :
How many 4-letter words, with or without meaning, can be formed out of the letters of the word, ‘GEEKSFORGEEKS’, if repetition of letters is not allowed ?

Explanation :
Total number of letters in the word ‘GEEKSFORGEEKS’ = 13
Therefore, the number of 4-letter words

```= Number of arrangements of 13 letters, taken 4 at a time.
= 13P4 ```

Example-2 :
How many 4-digit numbers are there with distinct digits ?

Explanation :
Total number of arrangements of ten digits ( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ) taking 4 at a time

`= 10P4 `

These arrangements also have those numbers which have 0 at thousand’s place.
(For eg- 0789 which is not a 4-digit number.).

If we fix 0 at the thousand’s place, we need to arrange the remaining 9 digits by taking 3 at a time.

Total number of such arrangements

`= 9P3  `

Thus, the total number of 4-digit numbers

`= 10P4  - 9P3`

Example-3 :
How many different words can be formed with the letters of the word “COMPUTER” so that the word begins with “C” ?

Explanation :
Since all the words must begin with C. So, we need to fix the C at the first place.
The remaining 7 letters can be arranged in 7P7 = 7! ways.

Example-4 :
In how many ways can 8 C++ developers and 6 Python Developers be arranged for a group photograph if the Python Developers are to sit on chairs in a row and the C++ developers are to stand in a row behind them ?

Explanation :
6 Python Developers can sit on chairs in a row in 6P6 = 6! ways
8 C++ Developers can stand behind in a row in 8P8 = 8! ways
Thus, the total number of ways

`= 6! x 8! ways `

Example-5 :
Prove that 0! = 1.

Explanation :
Using the formula of Permutation-

```P(n, r) = n! / (n-r)!

P(n, n) = n! / 0!  (Let r = n )

n! = n! / 0!    (Since, P(n, n) = n!)
0! = n! / n!
0! = 1
Thus, Proved ```

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