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  4. Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, do the word start with P?

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Q. Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, do the word start with $P$?

Permutations and Combinations

Solution:

There are $12$ letters, of which $N$ appears $3$ times, $E$ appears $4$ times and $D$ appears $2$ times and the rest all are different.
Therefore, the required number of arrangements $= \frac{12!}{3! 4! 2!} = 1663200.$
Let us fix $P$ at the extreme left position, we then count the arrangements of the remaining $11$ letters. Therefore, the required number of words starting with $P$ are
$=\frac{11!}{ 3! 2! 4!}= 138600.$