# Problem about combinatorics

How many ways are there to make seven $a$, eight $b$, three $c$, six $d$ in one row, so that there are not two pairs $cc$ AND $ca$ in ways?

My attempt:
I consider: cc,ca
All cases are: $${24!} \over {7!8!3!6!}$$ My answer is: $${{24!} \over {7!8!3!6!}}-k. \\ k={22! \over 6!7!8!}$$

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Your statement for all cases is correct. If you want the number of permutations without a series ccca, ccac, cacc, there are many more than 3 of them. The number with one of these three is $3($which of the three$)\cdot 20$(which position the set of four starts in$) \cdot$(the number of ways to order the other $20$ letters). You can work out the last-it is the same idea as your first. Not that this is simplified because we used all the c's in out set of four, so we didn't have to worry about double counting.

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Do you consider set of four ? we shouldn't use pairs ca AND cc in cases.can you explain more? Thanks. –  geni Dec 15 '12 at 17:00
@I edited my answer , does new answer is true? –  geni Dec 15 '12 at 17:18
@geni: in your original question it seemed the three c's and one a had to be in a block of 4. Now it seems the two pairs can be separate. You are right that you can then just consider them to be single characters, different from the rest, and I agree with your answer. It would be better to use a different character from ? and to explain how you got it. –  Ross Millikan Dec 15 '12 at 17:49