Question 658540
 I assume, for ordered combinations (i.e. permutations).

Removing the duplicate {{{M}}},{{{ A}}} and {{{T}}}, we have {{{8}}} different letters from which we can make 

[8]P[4]={{{ 8!/4!= 1680}}} permutations.

We need to {{{add}}} words having two {{{M}}}'s and two other distinct letters:

Pick two spots for the {{{M}}}'s ([4]C[2] ways) and then assign the {{{7}}} other letters the two other spots ([7]P[2]). 

So, we have ([4]C[2])([7]P[2]) = {{{6 * 42 = 252}}} more possibilities.

We do the same for words with two {{{A}}}'s and those with two {{{T}}}'s, so we're now at {{{1680 + 3*252 = 2436}}}.

We also need to {{{add}}} words having pairs of {{{identical}}}{{{ letters}}} like MAMA and TAAT... 

For two {{{M}}}'s and two {{{A}}}'s, there are [4]C[2]={{{6}}} possibilities. Same for {{{M}}}'s and {{{T}}}'s, and same for {{{A}}}'s and {{{T}}}'s. 
This add {{{18}}} to our total.

so, we have
{{{2436+18=2454}}}

Answer: {{{2454}}}