how many ways can the letters in the word SEVENTEEN be arranged so that E is at the middle?

Question

I have argued like this :

The number of arrangements with no restrictions 9!/(4!2!) The number of ways with NO E at the middle: 8!/(4!2!)

Then subtracting the two I get 840 ways.

Am I right?

Answer 1

I don't know how you got $\frac{8!}{4!2!}$, but no, you're not right.

Just reserve an E for the middle letter, and freely arrange the other 8 letters into the other 8 spots. There are $\frac{8!}{3!2!} = 3360$ ways to do this.

Answer 2

The count for distinct permutations of $\sf \underset 9{\underbrace{STV\overset 2{\overbrace{NN}}\overset 4{\overbrace{EEEE}}}}$ is indeed $9!/2!4!$

Now pick any $\sf E$ and glue it down in the center. Count the distinct permutations for the remaining letters (Ie $\sf STVNNEEE$) in the remaining places.