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August 20th, 2011, 09:43 AM  #1 
Newbie Joined: Feb 2008 Posts: 24 Thanks: 0  success arrangement
in how many letters of SUCCESS can be arranged such that no two C or no two S are together.

August 21st, 2011, 04:49 AM  #2 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  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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