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20 Mar 2009, 16:09
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Let each different arrangement of all the letters of DELETED be called a word. In how many of these words will the D's be separated?
I can get the number by first finding the total number of words which is 7!/(3!*2!) = 420 and then subtracting from it the total number of words that has D's together (this number is 6!/3! = 120). Can someone explain how I can find the result through regular counting process?
I can get the number by first finding the total number of words which is 7!/(3!*2!) = 420 and then subtracting from it the total number of words that has D's together (this number is 6!/3! = 120). Can someone explain how I can find the result through regular counting process?
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27 Mar 2009, 11:24
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420 is correct.
now if we take DD as one entity then we have 6! ways, but within this D is repeating so we get 6!/2! ...why do we take 6!/3! . Please explain.
now if we take DD as one entity then we have 6! ways, but within this D is repeating so we get 6!/2! ...why do we take 6!/3! . Please explain.
sanjay_gmat
27 Mar 2009, 14:03
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Well, the six elements in second case (both D's together) are :
1 - DD
2 - E
3 - E
4 - E
5 - L
6 - T
in this group, there are 3 repetitive elements, hence total number of permutations = 6!/3!.
I don't think you need to divide by 2! because the 2Ds are considered one single group.
