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.

Part (a) is $~\displaystyle \binom{15}{5}$.  That is, for each collection of $5$ distinct letters, there is only one way of ordering the letters (i.e. the letters must be in alphabetical order).
Part (b) is tricky, and permits two distinct approaches:

*

*the direct approach, where you examine each of the possible consonant first letters and consonant last letters possible.

*the indirect approach, which involves identifying the number of ways of violating the constraint, and then deducting that from the answer in part (a).

I prefer the 2nd (indirect) approach above, which might well be the problem composer's intent.  It allows you to use the answer in part (a) as a stepping stone.

The first thing to do is identify the location of each of the vowels:

*

*A : position $1$.

*E : position $5$.

*I : position $9$.

*O : position $15$.

The strategy will be to start with the part (a) answer, deduct all of the ways that a $5$ letter word can start with a vowel, deduct all of the ways that a $5$ letter word can end with a vowel, and then add back all of the ways that a word can both start and end with a vowel.

For a vowel in position $n ~: n \leq 11$, there are $~\displaystyle \binom{15 - n}{4}$ ways of selecting $4$ letters that follow the vowel.
So, let
$$A_1 = \binom{14}{4} + \binom{10}{4} + \binom{6}{4}.$$
For a vowel in position $n ~: n \geq 5$, there are $~\displaystyle \binom{n - 1}{4}$ ways of selecting $4$ letters that precede the vowel.
So, let
$$A_2 = \binom{4}{4} + \binom{8}{4} + \binom{14}{4}.$$
So, at this point, the running total is
$$\binom{15}{5} - (A_1 + A_2)$$
and it only remains to add back the number of ways of starting and ending with a vowel.

Here, I think it is best to dispense with elegance, take off my shoes, and count.
AxxxE : $~\displaystyle B_1 = \binom{3}{3}.$
AxxxI : $~\displaystyle B_2 = \binom{7}{3}.$
AxxxO : $~\displaystyle B_3 = \binom{13}{3}.$
ExxxI : $~\displaystyle B_4 = \binom{3}{3}.$
ExxxO : $~\displaystyle B_5 = \binom{9}{3}.$
IxxxO : $~\displaystyle B_6 = \binom{5}{3}.$

Final computation:
$$\binom{15}{5} - (A_1 + A_2) + (B_1 + \cdots + B_6).$$
