# 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. =^{13}P_{4 }

**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

=^{10}P_{4}

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

=^{9}P_{3}

Thus, the total number of 4-digit numbers

=^{10}P_{4 }-^{9}P_{3}

**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 ^{7}P_{7 } = 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 ^{6}P_{6 } = 6! ways

8 C++ Developers can stand behind in a row in ^{8}P_{8} = 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