Number of permutations of AABBBCC, taking 7 letters at a time, when repititions are allowed

Question

What is the number of permutations of the word AABBBCC, taking 7 letters at a time, repetitions being allowed?

I think it should be $3^7$, but I can't see why. Also what would be the number of permutations when only 5 letters are taken at a time, repetitions being allowed as above?

I came up with a general formula, $(Number\ of\ Different\ letters)^{Number\ of\ spaces\ to\ be\ filled}$

EDIT: One more confusion, what if, say we have p number of A's, q number of B's, r number of C's and s number of D's, and only three letter words are to be formed? The solution to one such problem in a book uses different cases, viz. all same, all different, etc., to count different number of arrangements, but it was only for 2 letters and (therefore), easy to do.

Answer 1

There you go:

• For the 2 A characters, choose 2 out of the 7 available places.
• For the 3 B characters, choose 3 out of the 5 remaining places.
• For the 2 C characters, choose 2 out of the 2 remaining places.

And the result is: $$\binom{7}{2}\times\binom{5}{3}\times\binom{2}{2}=\frac{7!5!2}{(2!5!)(3!2!)(2!0!)}=210$$