Question

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

Answer 1

Answer:60 I think

Step-by-step explanation:

Answer 2

Answer:

$18$.

Step-by-step 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 $(2 \times 3) \times (3 \times 2 \times 1) = 36$ 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:

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

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