# how many ways can the letters in ARRANGEMENT can be arranged

In how many different ways can the letters in the word ARRANGEMENT be arranged?

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In general if you have $n$ objects with $r_1$ objects of one kind, $r_2$ objects of another,...,and $r_k$ objects of the $k$th kind, they can be arranged in $$\frac{n!}{(r_1!)(r_2!)\dots(r_k!)}$$ ways. –  user48012 Nov 13 '12 at 15:45
@S.M. +1 I'd upvote it as an answer if you post it as an answer. It's always nice to see how problems of these kinds, in general, can be approached. –  amWhy Nov 13 '12 at 15:52
Nah, it is just a comment. –  user48012 Nov 13 '12 at 15:57

"ARRANGEMENT" is an eleven-letter word.

If there were no repeating letters, the answer would simply be $11!=39916800$.

However, since there are repeating letters, we have to divide to remove the duplicates accordingly. There are 2 As, 2 Rs, 2 Ns, 2 Es

Therefore, there are $\frac{11!}{2!\cdot2!\cdot2!\cdot2!}=2494800$ ways of arranging it.

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what is the proability that an arrangement chosen at random begins with the letters EE? –  darshanie M Nov 13 '12 at 19:25

The word ARRANGEMENT has $11$ letters, not all of them distinct. Imagine that they are written on little Scrabble squares. And suppose we have $11$ consecutive slots into which to put these squares.

There are $\dbinom{11}{2}$ ways to choose the slots where the two A's will go. For each of these ways, there are $\dbinom{9}{2}$ ways to decide where the two R's will go. For every decision about the A's and R's, there are $\dbinom{7}{2}$ ways to decide where the N's will go. Similarly, there are now $\dbinom{5}{2}$ ways to decide where the E's will go. That leaves $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a total of $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$

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what is the proability that an arrangement chosen at random begins with EE? –  darshanie M Nov 13 '12 at 19:27