How many $5$ letter words can be made from $15$ letter set where multiple conditions must be met a) How many $5$-letter words can be made using letters from the $15$ letter set $\{A, B, C ... , O\}$ such that the letters are all different and in alphabetical order?
b) How many are there if we add the condition that no word begins OR ends with a vowel?
I understand part a). It's just  $\binom{15}{5}$. But I am having trouble with b) I thought of creating two sets such as $A$ for all words that start with a vowel and set $B$ for all words that end in a vowel and then finding $A \cup B$ and subtract that from $\binom{15}{5}$ but I am not sure. Any help and guidance would be appreciated.

Here is simplification. Note that if a five letter word in alphabetical order contains letter $A$, it must start with $A$ and if it contains letter $O$, it must end with $O$. But as we cannot have a vowel at either end, that leaves us to make words with remaining thirteen letters,
B C D  E  F G H  I  J K L M N
If the set of words starting with a vowel is $P$ and the set of words ending with a vowel is $Q$,
$|P \cup Q| = |P| + |Q| - |P \cap Q|$
$ \displaystyle |P| = {9 \choose 4} + {5 \choose 4} = 131$
$ \displaystyle |Q| = {7 \choose 4} = 35$
$ \displaystyle |P \cap Q| = 1$
So the answer is $~ \displaystyle {13 \choose 5} - \left(131 + 35 - 1\right) = 1122$
