# Number of $3$ letters words from $\bf{PROPOSAL}$ in which vowels are in the middle

The number of different words of three letters which can be formed from the word $\bf{"PROPOSAL"}$, if a vowel is always in the middle are?

My try: We have $3$ vowels $\bf{O,O,A}$ and $5$ consonants $\bf{P,P,R,S,L}$. Now we have to form $3$ letter words in which vowels are in the middle.

First we will select $2$ consonants from $5$ consonants. This can be done in $\dbinom{5}{2}$ ways. The middle vowel can be selected in $\dbinom{3}{1}$ ways. So the total number of ways of selecting $3$ letter words is $\dbinom{5}{2}\cdot \dbinom{3}{1}$.

Now I did not understand how can I solve after that.

Help required. Thanks

• First of all, you need to account for the fact that $O$ appears twice, as does $P$: So $PAP$ and $PAP$ would count twice, instead of once (it should only count once). Second, there's nothing ruling out the possibility that the three letter word is $OOA$, $OAO$, or $PAO$, etc. If the problem stated "three letter words where vowels must always appear [only] in the middle... then you can assume consonants must appear on the ends. But without the word only, that assumption is unwarranted. Commented Mar 16, 2014 at 14:36

We have $3$ slots.$$---$$ The middle one has to be a vowel. There are only two ways in which it can be filled: O, A. Let us put O in the middle.$$-\rm O-$$ Consider the remaining letters: O, A, $\bf P_1$, $\bf P_2$, R, S, L. Since we will be getting repeated words when we use $\bf P_1$ or $\bf P_2$ once in the word, we consider both these letters as a single letter P. So, the letters now are O, A, P, R, S, L. We have to select any two of them (this can be done in $^6C_2$ ways) and arrange them (this can be done in $2!$ ways). So, total such words formed are $(^6C_2)(2!)=30$. Now there will be one word (P O P) where we will need both the P's. So we add that word. Total words=$31$.
Now, we put A in between.$$-\rm A-$$
Again proceeding as above, we have the letters O, P, R, S, L. Total words formed from them are $(^5C_2)(2!)=20$. We add two words (O A O) and (P A P). Thus, total words formed here=$22$
Hence, in all, total words that can be formed are $22+31=53$.