Question 417988
<pre><font face = "consolas" color = "indigo" size = 4><b>
If we could distinguish all 6 letters we would have 7! or 5040.
If we could not distinguish the 3 M's but could distinguish the 2 A's,
then since there are 3 positions in which the M's could be placed, this 
is 3! times too many, So we would divide that by 3! and get 840 ways, if 
the M's are not distinguishable but the A's are.
But since the A's are also indistinguishable, we must also divide that
by 2!.  So there are

6!/(3!2!) = 720/(6*2) = 720/12 = 60 ways. Here they are: 
  
 1. MAMMAL
 2. MAMMLA
 3. MAMAML
 <font color = "red">4. MAMALM</font>
 5. MAMLMA
 <font color = "red">6. MAMLAM</font>
 7. MAAMML
 <font color = "red">8. MAAMLM</font>
 <font color = "red">9. MAALMM</font>
10. MALMMA
<font color = "red">11. MALMAM</font>
<font color = "red">12. MALAMM</font>
13. MMAMAL
14. MMAMLA
15. MMAAML
<font color = "red">16. MMAALM</font>
17. MMALMA
<font color = "red">18. MMALAM</font>
19. MMMAAL
20. MMMALA
21. MMMLAA
22. MMLAMA
<font color = "red">23. MMLAAM</font>
24. MMLMAA
25. MLAMMA
<font color = "red">26. MLAMAM</font>
<font color = "red">27. MLAAMM</font>
28. MLMAMA
<font color = "red">29. MLMAAM</font>
30. MLMMAA
31. AMMMAL
32. AMMMLA
33. AMMAML
34. AMMALM
35. AMMLMA
36. AMMLAM
37. AMAMML
38. AMAMLM
39. AMALMM
40. AMLMMA
41. AMLMAM
42. AMLAMM
43. AAMMML
44. AAMMLM
45. AAMLMM
46. AALMMM
47. ALMMMA
48. ALMMAM
49. ALMAMM
50. ALAMMM
51. LMAMMA
52. LMAMAM
53. LMAAMM
54. LMMAMA
55. LMMAAM
56. LMMMAA
57. LAMMMA
58. LAMMAM
59. LAMAMM
60. LAAMMM

If we require that M's come first and last:

M _ _ _ _ M

Then we just have to fill in the middle 4 blanks with A,M,A,L,
using the same reasoning as above:

So the answer is 4!/2! = 24/2 = 12 ways. They are indicated in
red in the above list.

Edwin</pre>