How many $5$ letter words can be made from $15$ letter set where multiple conditions must be met

Question

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.

Answer 1

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 $~ \displaystyle {13 \choose 5} - \left(131 + 35 - 1\right) = 1122$

Answer 2

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.

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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.

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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.

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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 : $~\displaystyle B_6 = \binom{5}{3}.$

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Final computation:

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