# Permutations Alphabet

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!?

• b) $\frac{25!*10}{16!}$ c) $\frac{24!*9*10}{17!}$ ? Commented Nov 27, 2018 at 15:11
• Ok is this correct now? Commented Nov 27, 2018 at 15:16
• c) $\frac{24!*9*10}{16!}$ Commented Nov 27, 2018 at 15:18
• Yes @mathnoob that's correct, without any doubt Commented Nov 27, 2018 at 15:20
• For b) first I get all permutations of $9$ letters not including $z$, then I have to insert $z$ into the permutations, there are $10$ places to put $z$, hence multiply by $10$. For c) the reasoning is the same. Commented Nov 27, 2018 at 15:35

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.

• I have been using the permutations formula p(n,r) = n!/(n-r)! So I am not quite sure why one would use combinations? I thought this was a permutation question because the order matters.
– user618796
Commented Nov 27, 2018 at 15:17
• For b) first you choose $9$ letters out of $25$ because you already choose $z$. Then you permute $10$ letters so you multiply by $10!$. For c) first you choose $8$ letters out of $24$ because you already choose $a$ and $z$, then you permute $10$ letters. Commented Nov 27, 2018 at 15:21
• Yes, elaborate it a little , it will help other to understand Commented Nov 27, 2018 at 15:30
• Please check the elaborated answer and let me know if more is to be added. Commented Nov 27, 2018 at 15:31
• Yes man of course. Commented Nov 27, 2018 at 15:33