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August 19th, 2007, 02:23 AM   #1
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Permutation problem

Q: In how many ways the letters in the word PARALLEL can be arranged so that there will always be two "L"s together as in the original word?
And my solution is :
7!/2!-6!/2!
PARALLEL has 8 letters with one letter appearing 3 times and 1 letter appearing 2 times.
The letters can be arranged in 8!/(3!*2!) = 3360 ways.

If you want 2 Ls together, you can treat it is 1 single letter.

So it seems that the letters can be arranged in 7!/2!(= 2520) ways. but when 3 Ls are together we can't distinguish between the pair LL and lone L. So permutations 7!/2! (= 2520) include the permutation like "PAR(LL)LAE" & "PARL(LL)AE". This problem arise when 3 Ls are together. SO In how many permutations are they together? In 6!/2! permutations 3 Ls are together. But when you disregard the occasional identical character between LL and L, the number of times 3Ls apear together is doubled (we have disregarded this character to get 7!/2!). So 2*6!/2! number times 3 Ls have appeared together in our permutation (7!/2!), half of them are identical. So the number of valid permutation is 7!/2!-6!/2! = 2160

I will ask for vote. Whether you think I'm correct or wrong!
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August 19th, 2007, 01:45 PM   #2
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There are 6 positions for "LLL", 25 positions for "LL" at one end and another "L" separate from the "LL", and 54 position for "LL" elsewhere and another "L" separate from the "LL". That's a total of 36 ways of placing the "L"s with at least two together somewhere. For each of those ways, the "A"s can be added in 54/2! ways, i.e., 10 ways, and then the "P", "R" and "E" can be added in 3! ways, i.e., 6 ways. Hence the required total is 36106, i.e., 2160.
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August 20th, 2007, 08:14 AM   #3
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