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 .
how many ways can the letters in ARRANGEMENT can be arranged
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0possible duplicate of [How many different words can be formed using all the letters of the word GOOGOLPLEX?](http://math.stackexchange.com/questions/483277/how-many-different-words-can-be-formed-using-all-the-letters-of-the-word-googolp) – 2015-03-13
3 Answers
"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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1@darshanieM 0.072 – 2014-05-11
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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1@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. – 2016-01-02
11! = 39916800
We don't have to divide, add or remove anything, because it doesn't matter if some of the letters appear twice. It is not specified that the word ARRANGEMENT must fulfill the properties of a set
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0what is the proability that an arrangement chosen at random begins with the letters EE? – 2012-11-13