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03 Dec 2012, 12:02
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
95% (01:09) correct
5%
(01:30)
wrong
based on 246
sessions
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A radio station has to choose three days of the seven in a week to broadcast a certain program, and that set will repeat each week. The program can be broadcast equally on any of the seven weekdays —- weekdays vs. weekends don’t matter at all —- nor does it matter whether the days the program airs are adjacent or not. Absolutely any three of the seven weekdays can be chosen. How many different three-day combinations of the seven weekdays can be constructed?
(A) 9
(B) 15
(C) 21
(D) 35
(E) 56
For tips on calculating the answer to questions like this without a calculator, see this post:
https://magoosh.com/gmat/2012/gmat-math- ... binations/
Experts, feel free to share you own non-calculator tips for calculating nCr values.
Mike
(A) 9
(B) 15
(C) 21
(D) 35
(E) 56
For tips on calculating the answer to questions like this without a calculator, see this post:
https://magoosh.com/gmat/2012/gmat-math- ... binations/
Experts, feel free to share you own non-calculator tips for calculating nCr values.
Mike
04 Dec 2012, 00:59
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mikemcgarry
Pick 3 days out of 7.. 7C3 = 7!/(4!*3!) = 35
Ans D it is.
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27 Dec 2012, 02:38
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I used the ANAGRAM technique to answer this question:
D1 D2 D3 D4 D5 D6 D7
N N N Y Y Y N
ANAGRAM gets the permutation of the options and then divide it by the number of selected and the number of not selected. In our case, we are looking for the permutation of 7 days divided by 3 selections and 4 non-selections.
\(=\frac{7!}{4!3!}=7*5=35\)
Answer: D
D1 D2 D3 D4 D5 D6 D7
N N N Y Y Y N
ANAGRAM gets the permutation of the options and then divide it by the number of selected and the number of not selected. In our case, we are looking for the permutation of 7 days divided by 3 selections and 4 non-selections.
\(=\frac{7!}{4!3!}=7*5=35\)
Answer: D
