How many distinguishable permutations for the letters in the word "reassessed" are there?

"reassessed" has a total of 10 characters, one of which (e) occurs 3 times and another (s) which occurs 4 times.

Taking each character to be distinct, there would be a total of 10! permutations. But we don't want to do that, and instead want, for instance,

rEassessEd

and

rEassessEd

to count as the same permutation. So we divide the previous total by the number of ways we can permute each set of repeated characters. For example, 3 e's can be rearranged in 3! ways.

So the total number of distinguishable permutations would be

10!/(3! × 4!) = 25,200

1. d.90

2. e. none of these (80)

3. a. 0

4. b. 6

5. e. none of these (10)

6. a. 5