Seven letters are selected from a bag w/o replacement. What's the probability the selection contains the letters "MAT" in any order? Seven letters are selected without replacement from a bag containing 9 As, 2 Ms, 6 Ts, and 81 other letters. Each of the $98 \choose 7$ selections are equally likely. What is the probability that the selection contains the letters "MAT" in any order? 

Right off the bat, I thought it would be like choosing 1 of each desired letter, multiplied by the remaining letters in the bag that are none of the above:
$$\frac{(9)(2)(6)(81)(80)(79)(78)}{98 \choose 7}$$
Is this correct or am I missing something?
Also, is there a way to solve this using the Multinomial Theorem? Thanks in advance!
 A: Imagine that the letters are written on tiles, as in the game Scrabble.  Then we are selecting seven of the $98$ tiles in the bag.  We are also told that each of the 
$$\binom{98}{7}$$
selections are equally likely. Notice that those $\binom{98}{7}$ selections are not ordered.  Therefore, the number of favorable cases must not be ordered.  Consequently, we wish to answer the following question:

What is the probability that when seven letters are selected without replacement from a bag containing nine A's, six T's, two M's, and $81$ other letters that the selection contains at least one A, at least one M, and at least one T?

We can find the number of favorable cases by subtracting the number of selections in which there are no A's or no M's or no T's from the total number of selections.
Let $S_A$, $S_M$, and $S_T$ denote the set of selections of seven tiles that include, respectively, no A's, no M's, and no T's.  Then the unfavorable selections are those in $S_A \cup S_M \cup S_T$.  By the Inclusion-Exclusion Principle, the number of unfavorable selections is 
$$|S_A \cup S_M \cup S_T| = |S_A| + |S_M| + |S_T| - |S_A \cap S_M| - |S_A \cap S_T| - |S_M \cap S_T| + |S_A \cap S_M \cap S_T|$$ 
$|S_A|$:  Since there are nine tiles that display an A, $98 - 9 = 89$ tiles do not display an A.  Thus, there are 
$$\binom{89}{7}$$
ways to select seven tiles that do not include an A. 
$|S_M|$:  Since there are two tiles that display an M, $98 - 2 = 96$ tiles do not display an M.  Thus, there are
$$\binom{96}{7}$$
ways to select seven tiles that do not include an M. 
$|S_T|$:  Since there are six tiles that display an M, $98 - 6 = 92$ tiles do not display a T.  Thus, there are
$$\binom{92}{7}$$
ways to select seven tiles that do not include a T.
$|S_A \cap S_M|$:  Since there are nine tiles that display an A and two tiles that display an M, there are $98 - 9 - 2 = 87$ tiles that display neither an A nor a T.  Thus, there are
$$\binom{87}{7}$$
ways to select seven tiles that include neither an A nor an M.
$|S_A \cap S_T|$:  Since there are nine tiles that display an A and six tiles that display a T, there are $98 - 9 - 6 = 83$ tiles that display neither an A nor a T.  Thus, there are
$$\binom{83}{7}$$
ways to select seven tiles that include neither an A nor a T.
$|S_M \cap S_T|$:  Since there are two tiles that display an M and six tiles that display a T, there are $98 - 2 - 6 = 90$ tiles that display neither an M nor a T.  Thus, there are
$$\binom{90}{7}$$
ways to select seven tiles that include neither an M nor a T.
$|S_A \cap S_M \cap S_T|$:  Since there are nine tiles that display an A, two tiles that display an M, and six tiles that display a T, there are $98 - 9 - 2 - 6 = 81$ tiles that display none of those letters.  Thus, there are
$$\binom{81}{7}$$
ways to select seven tiles that include neither an A, nor an M, nor a T.
Hence, the number of unfavorable cases is
$$\binom{89}{7} + \binom{96}{7} + \binom{92}{7} - \binom{87}{7} - \binom{83}{7} - \binom{90}{7} + \binom{81}{7}$$
Therefore, the number of favorable cases is
$$\binom{98}{7} - \binom{89}{7} - \binom{96}{7} - \binom{92}{7} + \binom{87}{7} + \binom{83}{7} + \binom{90}{7} - \binom{81}{7}$$
Dividing that number by 
$$\binom{98}{7}$$ 
yields the probability that a selection of seven tiles from the bag includes at least one A, at least one M, and at least one T.
