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: ${{24!} \over {7!8!3!6!}}-k. \\ k={22! \over 6!7!8!}$