Question

Different words being formed by arranging the letters of the word "INTERMEDIATE". All the words obtained are written in the form of a dictionary,

Number of permutations if the words start with E & ends with I

• A. $907200$
• B. $\frac{10!}{2!}$
• C. $3628800$
• D. None of these

Answer 1

Answer: (A)

$907200$

We fix $E$ and $I$ at the beginning and the end respectively.

$E----------I$

The remaining letters are:

$NTERMDAT$

Thus, total number of permutations = $\frac{10!}{2!2!}=907200$ (we have 10 letters in between I and E...and in which 2E,2T are...there)

Hence, $\left(a\right)$ is correct.