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 4. MAMALM 5. MAMLMA 6. MAMLAM 7. MAAMML 8. MAAMLM 9. MAALMM 10. MALMMA 11. MALMAM 12. MALAMM 13. MMAMAL 14. MMAMLA 15. MMAAML 16. MMAALM 17. MMALMA 18. MMALAM 19. MMMAAL 20. MMMALA 21. MMMLAA 22. MMLAMA 23. MMLAAM 24. MMLMAA 25. MLAMMA 26. MLAMAM 27. MLAAMM 28. MLMAMA 29. MLMAAM 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