Mathematics | Problems On Permutations | Set 1

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