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 .


closed as off-topic by Marc van Leeuwen, Przemysław Scherwentke, anomaly, Davide Giraudo, MPW Mar 13 '15 at 15:12

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    $\begingroup$ 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. $\endgroup$ – user48012 Nov 13 '12 at 15:45
  • $\begingroup$ @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. $\endgroup$ – Namaste Nov 13 '12 at 15:52
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    $\begingroup$ Nah, it is just a comment. $\endgroup$ – user48012 Nov 13 '12 at 15:57
  • $\begingroup$ possible duplicate of How many different words can be formed using all the letters of the word GOOGOLPLEX? $\endgroup$ – Marc van Leeuwen Mar 13 '15 at 12:53

"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.

  • $\begingroup$ what is the proability that an arrangement chosen at random begins with the letters EE? $\endgroup$ – darshanie M Nov 13 '12 at 19:25
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    $\begingroup$ @darshanieM 0.072 $\endgroup$ – Shaurya Gupta May 11 '14 at 12:35

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 $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$

  • $\begingroup$ what is the proability that an arrangement chosen at random begins with EE? $\endgroup$ – darshanie M Nov 13 '12 at 19:27
  • $\begingroup$ This is definitely the best answer! Talking about duplicates or repeating letters can only be more confusing for beginners... $\endgroup$ – nbro Jan 2 '16 at 14:57
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    $\begingroup$ @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. $\endgroup$ – André Nicolas Jan 2 '16 at 15:13

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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    $\begingroup$ Do you consider 'ARRANGEMENT' to be the same arrangement of the letters as 'ARRANGEMENT'? (Quick, which letters did I swap in the second version?) $\endgroup$ – Steven Stadnicki Nov 13 '12 at 16:48
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    $\begingroup$ (Or, on a much smaller scale, how many different arrangements of 'MOM' are there?) $\endgroup$ – Steven Stadnicki Nov 13 '12 at 16:49
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    $\begingroup$ Note the word "different" in the original post. That's an essential word. $\endgroup$ – Cameron Buie Nov 13 '12 at 17:19
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    $\begingroup$ @StevenStadnicki: I think this answer is given because "arrange" is used as a verb in the question, and swapping identical letters is seen as a nontrivial act, even though the result is not visibly different from the initial situation. Much like pumping water out of a leaking boat at the same rate as it is entering is a nontrivial action, even though the water pumped out cannot be distinguished from the water leaking in. But if you like wordplay, there are of course also many ways of arranging letters that make them end up elsewhere than on a horizontal line. $\endgroup$ – Marc van Leeuwen Nov 13 '12 at 17:28
  • $\begingroup$ what is the proability that an arrangement chosen at random begins with the letters EE? $\endgroup$ – darshanie M Nov 13 '12 at 19:28

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