Combinatorics alphabet If say I want to arrange the letters of the alphabet a,b,c,d,e,f such that e and f cannot be next to each other.
I would think the answer was $6\times4\times4\times3\times2$ as there are first 6 letters then 4 as e cannot be next to f.
Thanks.

The $6$ numbers without any restriction can be arranged in $6!$ ways.
If we put $e,f$  together, we can arrange the $6$ numbers in $2!(5!)$ ways, as $e,f$ can arranged in $2!$ ways.
So, the required number of combinations is $6!-2(5!)$ 
