The answer is 2454, apparently. I am getting 1446, however. Can anyone point out my error ?
We need to partition the solution into the mutually exclusive cases of choices of 4 letters where we have no repetitions (e.g. MATH), where we have one repetition (e,g, MATT) and where we have two repetitions (e.g. MMTT). Then we compute the corresponding number of perms of each case, and add, thus:
We have 8 unique letters, so we can make 8C4 selections with no repetitions.
We have 3 pairs of repetitions {MM, AA, TT} which gives us 3 * 7C2 choices with one repetition, since having chosen, say, MM, we need to select 2 more letters from the remaining 7 unique letters.
We have 3C2 choices with 2 repetitions taken from the set {MM, AA, TT}
So our total number of perms is:
(#choices of 4, no reps * 4!) + (#choices of 4, 1 rep * 4!/2!) + (#choices of 4, 2 reps * 4!/(2!2!) ) = (8C4 * 24) + (3 * 7C2 * 12) + (3C2 * 6) = 672 + 756 + 18 = 1446
Where did I go wrong ?