Question

Find the number of arrangements of the letters of the word when words begin with $I$ and end in $P$.

• A. $12400$
• B. $12420$
• C. $12620$
• D. $12600$

Answer 1

Answer: (D)

$12600$

Answer 2

Let us fix $I$ and $P$ at the extreme ends ( $I$ at the left end and $P$ at the right end). We are left with $10$ letters. Hence, the required number of arrangements $= \frac{10!}{3! 2! 4!} = 12600$