Number of ways to form a 3-letter word with repetition allowed? The additional rule is: no letter can be used more often than it appears in MILLENNIUM? (Which is pretty logical I guess)
MILLENNIUM = MM, II, LL, NN, E, U
My logic:
Case 1: Double letters + 1 distinct letters (choose pair from M, I, L, N + 1 from E, U)
$$\dbinom{4}{1} \dbinom{2}{1} \frac{3!}{2!} = 24$$
Case 2: All 3 distinct letters (permute 3 from M,I,L,E,N,U)
$$_6P_3 = 120$$
Sum = 24 + 120 = 144
Part (b): What about a four-letter word?
My logic:
Case 1: Double letters + Double letters (choose two from M, I, L, N)
$$\dbinom{4}{2} \frac{4!}{2! 2!} = 36$$
Case 2: Double letters + 2 distinct letters (choose 1 from M, I, L, N)
$$\dbinom{4}{1} \frac{4!}{2!} = 48$$
Case 3: All 4 distinct letters (permute 4 from M,I,L,E,N,U)
$$_6P_4 = 360$$
Sum = 36+48+360 = 444
However my answer is incorrect for both. Is there a mistake I made somewhere?
 A: For both parts, the problem is with case 2 with one double letter: The letters MM, II, LL, NN are being prevented from being used as a single.
As B.A. noted in the comments Part a) Case 2 should be
$$\displaystyle\binom{4}{1}\binom{5}{1}\dfrac{3!}{2!},$$
and Part b) Case 2 should be:
$$\displaystyle\binom{4}{1}\binom{5}{2}\dfrac{4!}{2!}$$
A: MILLENNIUM is composed of 6 symbols with 4 of them having multiplicity of 2 $=\{M^2, I^2, L^2, N^2, E^1, U^1\}$
(A) 3 Letter 'words':
Using X,Y,Z as placeholders, these will be permutations of either XXY or XYZ (a double and single, or three singles). In the first case, choose X from M,I,L,N, and Y from the 5 remaining letters.  In the second case choose all 3 different letters from M,I,L,N,E,U.
 $$\text{Count} =\,^3P_{2,1}\,^4C_1\,^5C_1+\,^3P_{1,1,1}\,^6C_3 \\= \frac{3!}{2!\,1!}\frac{4!}{1!\,3!}\frac{5!}{1!\,4!}+\frac{3!}{1!\,1!\,1!}\frac{6!}{3!\,3!} \\= 60+120=180$$
(A) 4 Letter 'words':
Using W,X,Y,Z as placeholders, these will be permutations of either WWXX, WWXY, WXYZ, with the placeholders filled as above.
$$\text{Count} =\,^4P_{2,2}\,^4C_2 + \,^4P_{2,1,1}\,^4C_1\,^5C_2 +\,^4P_{1,1,1,1}\,^6C_4 \\= \frac{4!}{2!\,2!}\frac{4!}{2!\,2!}+\frac{4!}{2!\,1!\,1!}\frac{4!}{1!\,3!}\frac{5!}{2!\,3!} + \frac{4!}{1!\,1!\,1!\,1!}\frac{6!}{4!\,2!} \\= 36 + 480 + 360\\= 876$$
A: To find the number of different ways to spell any word is simply counting up all the letters of that word, taking that total in factorial and dividing it by the factorial of all the duplicate letters. Ex. "Millennium" - 10!/(2!*2!*2!*2!) 
