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Director
Status: Matriculating
Affiliations: Chicago Booth
Joined: 03 Feb 2011
Posts: 947
Followers: 10
Kudos [?]:
144
[0], given: 123
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Question Stats:
25% (00:00) correct
75% (02:01) wrong based on 4 sessions
hey fluke - I am not able to find any tag that says - "permutation". So pls change the tag if required. In how many different ways can the letters A, A, B, B, B, C, D, E be arranged if the letter C must be to the right of the letter D? 1,680 2,160 2,520 3,240 3,360I dont have the OA for this one. But I believe answer is A. Pls verify the reasoning.
There are 8!/(2!*3!) ways to arrange the letters - A, A, B, B, B, C, D, E. Hence half of these arrangements will have C to the right of D and vice- versa.
Total ways = 8!/(2!*3!) / 2 = 8 * 5040 / 2 * 6 * 2 = 1680. Answer A
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Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2098
Followers: 109
Kudos [?]:
664
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gmat1220 wrote: hey fluke - I am not able to find any tag that says - "permutation". So pls change the tag if required. In how many different ways can the letters A, A, B, B, B, C, D, E be arranged if the letter C must be to the right of the letter D? 1,680 2,160 2,520 3,240 3,360I dont have the OA for this one. But I believe answer is A. Pls verify the reasoning.
There are 8!/(2!*3!) ways to arrange the letters - A, A, B, B, B, C, D, E. Hence half of these arrangements will have C to the right of D and vice- versa.
Total ways = 8!/(2!*3!) / 2 = 8 * 5040 / 2 * 6 * 2 = 1680. Answer A Voila!! Your reasoning is same as Bunuel's and you know what it means!!! You have definitely attained a good level of understanding of permutations and combinations. Good job!! Discussed here: probability-q-91460.html
_________________
~fluke
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Director
Status: Matriculating
Affiliations: Chicago Booth
Joined: 03 Feb 2011
Posts: 947
Followers: 10
Kudos [?]:
144
[0], given: 123
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Thanks fluke ! I love this forum. cheers
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