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.