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August 20th, 2011, 09:43 AM   #1
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success arrangement

in how many letters of SUCCESS can be arranged such that no two C or no two S are together.
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August 21st, 2011, 04:49 AM   #2
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Re: success arrangement

First, there are 7 letters in SUCCESS so if they were all differrent (imagine the two C being different colors and the three S being different colors!) there would be 7! arrangements. But there are 2 C so swapping them would give the same arrangement- 7! counts both so we need to divide by 2. Similarly, there are 3 S so the 3!= 6 ways or rearranging them only should not be counted- there way of arranging the letters in SUCCESS.

Now, to find the number of ways we can arrange them so that "no two C or no two S are together", we find the number of arrangements in which the are together and subtract. We can calculate the number of ways with the two C together by treating "CC" as a single letter. Then we have 6 letter word with 3 S. The number of ways of arranging 6 letters with 3 the same is [latex]\frac{6!}{3!}[/itex]. You want to subtract that off.

The number of ways you can arrange a word with 3 letters the same but no 2 of them are together seems much harder. I will have to think about that.
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