Using all the letters of the word ARRANGEMENT how many different words using all letters at a time can be made such that both A, both E, both R both N occur together .
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17In 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. – Nov 13 '12 at 15:45
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@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
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1Nah, it is just a comment. – Nov 13 '12 at 15:57
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possible duplicate of How many different words can be formed using all the letters of the word GOOGOLPLEX? – Marc van Leeuwen Mar 13 '15 at 12:53
2 Answers
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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 $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$
André Nicolas
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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
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2@darshanieM: By accident I saw your (by now very old) question. Apologies for not answering sooner, I may not have been pinged. To answer, it is cleanest to imagine the repeated letters are given ID numbers to make them distinct. There are then $11!$ equally likely words. For $EE$ at the beginning, we want to start with $E_1E_2$ or $E_2E_1$. For each there are $9!$ ways to finish. So the probability is $\frac{2\cdot 9!}{11!}$, that is, $2/55$. A simpler way: the probability the first letter was $E$ is $2/11$. Given first was $E$, probability second is $E$ is $1/10$. Multiply. – André Nicolas Jan 02 '16 at 15:13
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"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.
Or Svantovit
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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
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