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Difficult Math Problem #64 - Combinations

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by gmat_enthus » Mon Nov 27, 2006 9:48 pm
C

10!/(4!*3!*2!)
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destination 760! completed!
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Answer

by lalitaroral » Tue Nov 28, 2006 8:06 am
No doubt Its C
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OA

by 800guy » Wed Nov 29, 2006 2:41 pm
OA:

Total # of alphabets = 10
so ways to arrange them = 10!

Then there will be duplicates because 1st S is no different than 2nd S.
we have 4 Is
3 S
and 2 Ps

Hence # of arrangements = 10!/4!*3!*2!
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need some understanding on combinatorics

by ashish1354 » Sat Sep 20, 2008 3:22 am
i need to understand the logic behind dividing 10! by the repeat factorials of I, S & P can someone please explain??
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by cramya » Sat Sep 20, 2008 8:41 am
When u calculate 10! u treat all the letters to be different(Essentially u r saying if I take the first S and put it as starting letter of a word then thats a word and if I take the 2nd S and put it it as starting letter of a word its a different word etc.. But really both are the same words. Since there are 3 S's they could have been arranged in 3! ways You divide by the dupliactes to get to distinct words that could have been formed by these letters)

Similarly for P and I u do the same thing

Hope this helps!
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