I am not very sure about the methodology. Please let me know if the logic is faulty anywhere.
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 $22+31=53$.
(P.S: Please edit if anything wrong.)
If repetition is allowed:
- Choose $3$ places for vowels: $\binom{11}{3}$
- Fill those places with vowels: $5^3$
- Fill the remaining places with non-vowels: $(26-5)^{11-3}$
Hence the number of such sequences is $\binom{11}{3}\cdot5^3\cdot(26-5)^{11-3}$.
Best Answer
There are $\binom{8}{3}$ ways to choose the positions for the A's. Each of the five remaining positions can be filled with a B or a C. Hence, the number of such words is $$\binom{8}{3}\cdot 2^5$$
Consider a five-letter word composed of B's and C's such as $$BBCBC$$ We wish to insert three A's so that no two of them are consecutive. We have six spaces in which to insert the three A's, indicated by the squares in the example. $$\square B \square B \square C \square B \square C \square$$ To ensure that no two A's are consecutive, we must choose three of these six spaces into which to insert an A. For instance, if we choose the first, second, and fifth spaces, we obtain the sequence $$ABABCBAC$$
There are $2^5$ five-letter words composed of B's and C's and $\binom{6}{3}$ ways to choose three of the six spaces in which to insert an A. Hence, there are $$2^5 \cdot \binom{6}{3}$$ eight-letter words composed from the letters A, B, and C with exactly three A's in which no two of the A's are consecutive.