I would like to calculate thenumber of anagrams such that there aren't three consecutives that are the same. Yet I don't really know how to proceed any ideas ?


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    $\begingroup$ Can you compute the total number of anagrams? Now consider the A's as a block and compute the number. Subtract these. Also subtract the number with the P's as a block. The ones with both the A's and P's as a block.... $\endgroup$ – Ross Millikan May 29 '17 at 16:34
  • $\begingroup$ ...and add that last number ! $\endgroup$ – Evargalo May 29 '17 at 16:35

First find that there are $\frac{16!}{3!3!2!}$ anagrams, then subtract those having consecutive $A$'s. You can calculate their number pretending that the three $A$'s are a single letter. So $-\frac{14!}{3!2!}$. Do this again for the $P$'s. But this way you have subtracted twice the anagrams having both $A$'s and $P$'s grouped, that have to be added in back: $+\frac{12!}{2!}$. All in all: $$\frac{16!}{3!3!2!}-2\times\frac{14!}{3!2!}+\frac{12!}{2!}$$

  • $\begingroup$ Thank you, great explanation ! $\endgroup$ – J. OK May 29 '17 at 16:55

Since there's only two letters that are repeated $3$ times one can use the inlcusion/exclusion principle without too much hassle:

There's $\dfrac{16!}{3!3!2!}$ total anagrams, out of those $\dfrac{14!}{3!2!}$ contain the block $AAA$ and $\dfrac{14!}{3!2!}$ contain the block $PPP$, but we've double counted some, since $\dfrac{12!}{3!2!}$ contain both blocks.

The total amount is thus:

$$\dfrac{16!}{3!2!} - 2\dfrac{14!}{3!2!} + \dfrac{12!}{3!2!}$$

  • $\begingroup$ Great explanation ! $\endgroup$ – J. OK May 29 '17 at 16:55

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