Question

If the letters of the word mathematics are arranged arbitarily, the probability that c

Answer 1

You were headed in the right direction, but you went way too far. The C(35,7) is much too much. It counts, for example, "-M---I-T..." and "--M-I--T..." as different, even though both are basically "MIT..."

Here's a way to think of it. Starting from TTAA, allow yourself to insert the other letters, C, E, H, I, M, M, and S, one at a time, anywhere before, between, or after letters already in position. (It might help to picture the second M as, temporarily, an N, remembering to divide by 2 when you're done.) The C have 5 places it can go, the E then has 6, the H has 7, and so forth, for a total of

(5⋅6⋅7⋅8⋅9⋅10⋅11)/2

Can you now do the other examples you asked about (both A's before both M's or both M's before the E)? The first of those should actually be very easy!