Permutations Alphabet

Question

how many ways are there to arrange 10 letters taken from the alphabet a-z such that:

a) a is not included

b) z is included

c) both a and z are included

I believe for the first one it should be 25!/15!, but would both b and c be 26!/16!?

Answer 1

For second case it should be $25\choose 9$$×10!$ In this case , I have chosen $9$ letters out of $25$ since $Z$ is already included. After that I arranged the letters

For third case it should be $24\choose 8$$×10!$. In this case , I have chosen $8$ letters out of $24$ since $Z$ and $A$ are already included. After that I arranged the letters

Note: $n\choose r$ stands for combination.