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How do you solve this question.

I first solved this question by solving 9 factorial but the answer was wrong. Then I thought of doing it by taking in consideration about how many times the same letter appears but the whole thing just confuses me.

Pls help.

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    The reason the factorial ends up with the wrong answer is that it will not take into account that rearrangement of Ps doesn't make it another arrangement. Also Es are subject to this effect. I think this has been asked and aswered before, but I cannot find the old question. – 2017-02-24
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    Choose $3$ from $9$ places for the letter $P$, choose $2$ from the remaining $6$ places for the letter $E$ to get $n=\binom{9}{3}\binom{6}{2}4!$ – 2017-02-24
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    @PeterMelech Yes, that's another way to solve it and perhaps it deserves a separate answer. To be consequent one should perhaps write it $n=\binom93\binom62\binom41\binom31\binom21\binom11$? – 2017-02-24
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    @skyking Yes perhaps...anyway Your answer is sufficient. – 2017-02-24

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For the first we first look at how many ways you can arrange $P_1INE_1AP_2P_3LE_2$ which is $9!$, but this distinguishes between the three $P$s and the two $E$s. To correct for this fact we then consider one such arrangement and count how many ways one can rearrange the $P$s and $E$s that is how many arrangements that are equivalent which is $3!2!$. The number of arrangements are therefore ${9!\over2!3!}$.

The next is solved in similar way, but since we require it to start and end with $E$ we only have $7$ letters to arrange. The result therefore becomes $7!\over3!$. Here we don't correct for two $E$s since we placed them at the beginning and the end in only one way.

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    Don´t You mean two $E$ sin the 2nd line? a different approach leads to the same number (see my comment) – 2017-02-24
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    @PeterMelech Yes, I've changed that. – 2017-02-24