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

$=3\times 5\times 4$


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

$=3\times3 \times 4$


Total number of required words of four letters



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

  • $\begingroup$ 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. $\endgroup$
    – lulu
    Sep 15, 2018 at 11:41
  • 1
    $\begingroup$ 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 $12$. $\endgroup$
    – lulu
    Sep 15, 2018 at 11:46
  • $\begingroup$ @lulu: Yes , AAEM.. will work but not AAEE. If second case gives 12, I can't still get 120. $\endgroup$ Sep 15, 2018 at 11:53
  • 2
    $\begingroup$ 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. $\endgroup$
    – lulu
    Sep 15, 2018 at 11:57
  • $\begingroup$ @lulu: I have edited the question for clarification $\endgroup$ Sep 15, 2018 at 12:52

1 Answer 1


If we take two 'E' then rest three places can be filled by total =3×3×4 =36

(take two E)(another one Q other than A)(choose a position for Q) so

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

Why you get 36?


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