Answer:
[tex]18[/tex].
Step-by-step explanation:
Start by considering the two "[tex]\verb!P![/tex]"'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 [tex]\left(\genfrac{}{}{0}{}{2}{1}\right) = 2[/tex] times that of the actual value.
Start by fixing the first and the last letter since they come with special requirements.
There are [tex]2[/tex] choices for the first letter (the vowel: [tex]\verb!A![/tex] and [tex]\verb!E![/tex].)
Under the assumption that the two "[tex]\verb!P![/tex]"'s are distinct, there would be [tex]3[/tex] choices for the last letter ([tex]\verb!L![/tex] and the two
After choosing the first and the last letters, there would be [tex](5 - 1 - 1) = 3[/tex] choices for the second letter, [tex](3 - 1) = 2[/tex] choices for the third, and [tex](2 - 1) = 1[/tex] for the fourth.
Thus, under the assumption that the two "[tex]\verb!P![/tex]"'s are distinct, there would be [tex](2 \times 3) \times (3 \times 2 \times 1) = 36[/tex] ways to choose the new word. Since this assumption doubles the number of arrangements, dividing the [tex]36[/tex] by [tex]\left(\genfrac{}{}{0}{}{2}{1}\right) = 2[/tex] would give the actual number of arrangements:
[tex]\begin{aligned}\frac{36}{\genfrac{(}{)}{0}{}{2}{1}} = \frac{36}{2} = 18\end{aligned}[/tex].
[tex]\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}[/tex].
