Originally Posted by

**Elusive1324** 1. EDITED: I realize this doesn't come out to 319. I wonder what I'm missing ... can you post the entire problem statement?

Keep a running "list" of possible outcomes and analyze your possibilities based on what's in this list.

How many ways are there to select 1 letter from INSIPIDITY? There are 7 unique letters: I, N, S, P, D, T, Y and if you pick one at a time, then there are 7 possible selections. Put each possible outcome in your list: (I, N, S, P, D, T, Y)

How many ways are there to select 2 letters from INSIPIDITY? Again -- 7 unique letters.

Suppose you first pick I, how many possible selections are left? Since you can't re-pick I (because you don't count duplicates and don't replace the letter you've chosen), there are 6 selections left: (IN, IS, IP, ID, IT, IY). So if you can only select 2 unique letters at a time and your first pick was I, then there are 6 possibilities left. Put those 6 possibilities in your list: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY). What if you had picked N first? Then there are 6 selections possible: (NI, NS, NP, ND, NT, NY) -- add them to your list BUT notice that NI and IN are 'the same outcome' (since ordering doesn't matter). The list you now have should be: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY, NS, NP, ND, NT, NY). What if you had picked S first? Then, if we again list the possible outcomes AND remove the permutations we already have in the list you get: (SP, SD, ST, SY). Add those 4 outcomes to your list because you don't have them yet. What if you had picked P first? D? T? Y? (PD, PT, PY), (DT, DY), (TY).

Your list of outcomes should be: (I, N, S, P, D, T, Y, IN, IS, IP, ID, IT, IY, NS, NP, ND, NT, NY, SP, SD, ST, SY, PD, PT, PY, DT, DY, TY). Notice that 7 of those outcomes came from constraining yourself to select 1 unique letter and the other 21 came from constraining yourself to select 2 unique letters. Hopefully, you can see that by constraining ourselves to pick only 'k' unique letters from 7 possible unique letters, there is exactly "7 choose k" possible outcomes: 7 choose 1 = 7, 7 choose 2 = 21. Our list of possible outcomes for k=1 and k=2 is 7 + 21 = 28 possible outcomes.

How many unique ways are there to select 3 letters? 4? 5? 6? 7? 8? 9? (of course, beyond 7, you ran out of unique letters so n choose k, for k>n is always 0). How many outcomes is in your list after counting all of the unique ways? I count 127.