How many distinct four-letter words beginning with A can be formed from letters with two similar letters and two different letters?

Four-letter words are made from letters A, A, D, E, E, M, S, Y such that two letters are similar & another two are different & each word begins with letter 'A'. The total number of such words will be

a) $$\ 60 \quad$$ b) $$\ 80\quad$$ c) $$\ 100\quad$$ d) $$\ 120$$

My try:

Distinct letters: A, D, E, M, S, Y

Since each word begins with 'A' hence word structure will be $$\ \boxed{A} \boxed{X}\boxed{X}\boxed{X}$$

If we take another 'A' then rest three places can be filled by total

$$=3\times 5\times 4$$

$$=60$$

If we take two 'E' then rest three places can be filled by total

$$=3\times3 \times 4$$

$$=36$$

Total number of required words of four letters

$$=60+36$$

$$=96$$

but there is no option for $$96$$. My answer is wrong. My teacher says that option (d) 120 is correct answer. But I don't know how. Somebody please help me solve this problem. Thanks

• The phrasing is unclear. Does it mean that there is one duplicate, say $A,A$ and the other two are distinct? So...$AAEM$ would work but $AAEE$ would not? Or does it mean something else? We could reverse engineer the question from the official answer, but that's not a good way to do things.
– lulu
Commented Sep 15, 2018 at 11:41
• Note: I can't follow your second case computation. If the first letter is $A$ and there are two $E's$ then we have $3$ ways to place the two $E's$ and $4$ ways to fill the empty slot. That's $12$.
– lulu
Commented Sep 15, 2018 at 11:46
• @lulu: Yes , AAEM.. will work but not AAEE. If second case gives 12, I can't still get 120. Commented Sep 15, 2018 at 11:53
• As I say, the phrasing is very unclear. I get $72$ but that's using my interpretation of the question (and I probably have it wrong). That said, I don't see a way to get to $120$ (but perhaps I lack imagination). I'd ask for clarification on the question.
– lulu
Commented Sep 15, 2018 at 11:57
• @lulu: I have edited the question for clarification Commented Sep 15, 2018 at 12:52

$$(1)\cdot({|\{D,M,S,Y\}| \choose 1})\cdot({3\choose1})=4\cdot3=12.$$