Question

"How many ways are there to arrange some, but not all, of the letters in "Suspension" if at most once copy of each letter can be used?
I have a start... I think the part that says "at most" one copy means that you have to remove one of the "s"'s thus having 9 letters to work with (not 10). Then you have to start with case 1: that is one in the right spot. But case 1 also has the subsets of (no copies of s, 1 copy of s, no copies of n, and 1 copy of n). But what about no copy of s and 1 copy of n and no copy of n and 1 copy of s... Please advise. Note: that is only case 1 (where 1 letter is in the same spot). I guess I would have to go all the way up to 8 in the right spot "some but not all".

Answer 1

You have 7 letters in the given word:

S, U, P, E, N, I, O

Now, you have to arrange "some, but not all of these letters"

Number of ways to arrange 1 letter = 7P1
Number of ways to arrange 2 letters = 7P2

Where, P stands for the usual permutation

You have to continue this until 'ways to arrange 6 letters'. (You don't arrange 7 letters because you may not use all). Then, you add these up for the total number of ways.

Final answer = 7P1 + 7P2 + 7P3 + 7P4 + 7P5 + 7P6

Answer 2

Hmmmm.... can it really be that simple? WHat about the no copies of s, 1 copy of s, no copies of n, 1 copy of n?

Answer 3

Thanks Arora!
After more thought I believe all cases are covered in your eloquent solution : )