Given: We have 12 letters

To Find: Number of words formed with Letter of the word ‘PERMUTATIONS.

The formula used: The number of permutations of n objects, where p1 objects are of one kind, p2 are of the second kind, ..., pk is of a kth kind and the rest if any, are of a different kind is =\(\frac{n!}{p_1!p_2! ....p_k!}\)

In the word ‘PERMUTATIONS’ we have 2 T’s.

We have to start the word with P and end it with S, hence the first and last position is occupied with P and S respectively.

As two positions are occupied the remaining 10 positions are to be filled with 10 letters in which we have 2 T’s.

NOTE:- Unless specified , assume that repetition is not allowed.

Let us represent the arrangement

Hence,

The first place is occupied by P = 1 way

The last place (12th) is occupied by S = 1 way

For the remaining 10 places:

Using the above formula

Where,

n=10

p1=2

\(\frac{10!}{2!}\)= 1814400

Total number of ways 1x1814400x1 = 1814400 are ways.

In 1814400 ways the letters of the word ‘PERMUTATIONS’ can be arranged if each word starts with P and ends with S.