Question

How many permutations are there of the letters APPLE, if the arrangement must begin with a

vowel and end with a consonant?

Answer 1

`18`.

Explanation:

Start by considering the two "
`\verb!P!`"'s as being different from one another. That way, all five letters in this word would be distinct. With this assumption, the number of arrangements would be
`\left(\genfrac{}{}{0}{}{2}{1}\right) = 2` times that of the actual value.

Start by fixing the first and the last letter since they come with special requirements.

There are
`2` choices for the first letter (the vowel:
`\verb!A!` and
`\verb!E!`.)

Under the assumption that the two "
`\verb!P!`"'s are distinct, there would be
`3` choices for the last letter (
`\verb!L!` and the two

After choosing the first and the last letters, there would be
`(5 - 1 - 1) = 3` choices for the second letter,
`(3 - 1) = 2` choices for the third, and
`(2 - 1) = 1` for the fourth.

Thus, under the assumption that the two "
`\verb!P!`"'s are distinct, there would be
`(2 * 3) * (3 * 2 * 1) = 36` ways to choose the new word. Since this assumption doubles the number of arrangements, dividing the
`36` by
`\left(\genfrac{}{}{0}{}{2}{1}\right) = 2` would give the actual number of arrangements:

`\begin{aligned}\frac{36}{\genfrac{(}{)}{0}{}{2}{1}} = (36)/(2) = 18\end{aligned}`.

`\begin{aligned}& 1 && \verb!APPEL!\\& 2 && \verb!APLEP!\\& 3 && \verb!APEPL!\\& 4 && \verb!APELP!\\& 5 && \verb!ALPEP!\\& 6 && \verb!ALEPP!\\& 7 && \verb!AEPPL!\\& 8 && \verb!AEPLP!\\& 9 && \verb!AELPP!\\& 10 && \verb!EAPPL!\\& 11 && \verb!EAPLP!\\& 12 && \verb!EALPP!\\& 13 && \verb!EPAPL!\\& 14 && \verb!EPALP!\\& 15 && \verb!EPPAL!\\& 16 && \verb!EPLAP!\\& 17 && \verb!ELAPP!\\& 18 && \verb!ELPAP!\end{aligned}`.