Question

In how many ways can the letters of the word permutations be arranged if words starts starts with p and ends with s

Answer 1

hope this helps you

Answer 2

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There are 12 letters in given word , of which T occurs 2 times

Let us fix P and S at the extreme ends

Now , we are left with 10 letters

Thus , the required number of arrangement is

$= \frac{10!}{2!} \\ \\ = \frac{10 \times 9 \times 8 \times 7 \times 6 \times5 \times 4 \times 3 \times \cancel{ 2! }}{ \cancel{2!}} \\ \\ = 1814400$

Hence , the required number of permutations is 1814400