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Topic: Permutation Combination 1 (Read 148 times) 

navdeep1771
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Permutation Combination 1
« on: Jun 17^{th}, 2018, 12:06am » 
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In how many ways letters of the word 'ARRANGE' can be arranged such that "no two R's are together and no two A's are together"?


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towr
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Re: Permutation Combination 1
« Reply #1 on: Jun 17^{th}, 2018, 12:26pm » 
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I think it's 7!5!*2!*2!=4560 7! ways to rearrange 7 letters, minus 5! ways to rearrange the 5 distinct letters multiplied by twice 2 ways to arrange duplicate letters next to each other. That's assuming the duplicate letters are distinguishable. (Otherwise divide by 4)


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dudiobugtron
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Re: Permutation Combination 1
« Reply #2 on: Jun 18^{th}, 2018, 12:10am » 
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There are 5! ways to arrange ARNGE. Then for each of those, there are 4 places to put the second A, and then 5 places you can put the second R. So 5! x 5 x 4 = 2400 ways to arrange ARRANGE like that. This is different from towr's solution. I think this is because: towr's solution only subtracted the results with AA *and* RR, so for eg AARANGER would not have been subtracted.


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towr
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Re: Permutation Combination 1
« Reply #3 on: Jun 18^{th}, 2018, 8:53am » 
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Well, we can't both be right, but we can both be wrong. Counting them with a script says it should be 2640 There are 5 places to put the second A, because you can put the second R between the two A's if they're adjacent.


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towr
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Re: Permutation Combination 1
« Reply #4 on: Jun 18^{th}, 2018, 10:03am » 
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Fixing the mistake that dudiobugtron spotted: 7!  (2 * 6!*2!  5!*2!*2!) = 2640


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navdeep1771
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Re: Permutation Combination 1
« Reply #5 on: Jun 18^{th}, 2018, 9:03pm » 
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Duplicate letters are not distinguishable.

« Last Edit: Jun 18^{th}, 2018, 9:15pm by navdeep1771 » 
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navdeep1771
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Re: Permutation Combination 1
« Reply #7 on: Jun 19^{th}, 2018, 1:59am » 
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Yup. We have to divide by 2!.2!=4 And 660 is absolutely correct.


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dudiobugtron
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Re: Permutation Combination 1
« Reply #8 on: Jun 19^{th}, 2018, 11:58pm » 
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on Jun 18^{th}, 2018, 8:53am, towr wrote:Well, we can't both be right, but we can both be wrong. 
 lol. Thank you for the correction.


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Grimbal
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Re: Permutation Combination 1
« Reply #9 on: Jun 21^{st}, 2018, 1:43pm » 
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I had no time to write the solution at that time but this I quickly came up with an elegant solution: There are 7! permutations of the letters. There are 6! that contain 'AA' (just think of AA as a single symbol). Likewise, there are 6! permutations containing 'RR'. If you take 7!  6!  6! you remove all the wrong permutations, but you remove twice the combinations containing 'AA' and 'RR'. So you have to add that number back. It's 5!. So, the final result is 7!  6!  6! + 5! = 3720. Divide by 4 if letters are indistinguishable. 3720/4 = 930. Tadaaaa! ... er.... OK, the mistake is that 7! must be divided by 4 to take into account the duplicates. But 6! must be divided by 2. 5! counts no duplicates. So I get: 7!/4  6!/2  6!/2 + 5! = 660


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