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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.

Shubham
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    It's not $3^7$ -- that would be if you had unlimited amounts of each letter. In this case the standard formula would be $\frac{7!}{2! 3! 2!}$. The $7!$ on top is the total number of way to rearrange those letters if they were all distinct. However, they are not, so we divide by $2! 3! 2!$ since $A, B, C$ appears twice, thrice, and twice, respectively. – MT_ Apr 04 '14 at 11:53
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    Think about the problem this way: How many ways can you assign the three $B$'s to the various 7 spots available in a permutation? How many different ways can you assign the two $A$'s to the remaining 4 spots? You can only assign the $C$'s in one way to the last two spots. – Foo Barrigno Apr 04 '14 at 11:56
  • I think I wasn't able to frame the question. I meant to have Unlimited amounts of each letter, but then specifying that there are 2 A's, 3B's and 2C's doesn't make sense. Thanks, by the way! – Shubham Apr 04 '14 at 12:04
  • Can I use the method given in [link]http://math.stackexchange.com/questions/20238/6-letter-permutations-in-mississippi?rq=1 – Shubham Apr 04 '14 at 13:03

1 Answers1

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

barak manos
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