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Updated on: 19 Mar 2013, 09:21
Originally posted by testprep2010 on 28 Jun 2010, 10:48.
Last edited by Bunuel on 19 Mar 2013, 09:21, edited 1 time in total.
Last edited by Bunuel on 19 Mar 2013, 09:21, edited 1 time in total.
Renamed the topic and edited the OA.
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
70% (01:01) correct
30%
(01:38)
wrong
based on 652
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How many different 6-letter sequences are there that consist of 1A, 2B's and 3C's ?
A. 6
B. 60
C. 120
D. 360
E. 720
I thought the answer is 6!/2!3! = 60
But the answer is not 60.
A. 6
B. 60
C. 120
D. 360
E. 720
I thought the answer is 6!/2!3! = 60
But the answer is not 60.
28 Jun 2010, 11:02
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testprep2010
# of different permutations of 6 letters ABBCCC is indeed \(\frac{6!}{2!3!}=60\), so you are right answer must be B (60).
THEORY:
Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).
For example number of permutation of the letters of the word "gmatclub" is \(8!\) as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
General Discussion
