A radio station has to choose three days of the seven in a week to broadcast a certain program, and that set will repeat

[Vincen]

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

[spoiler]OA=D[/spoiler]

Source: Magoosh

[Jay@ManhattanReview]

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

[spoiler]OA=D[/spoiler]

Source: Magoosh

The no. of different three-day combinations of the seven weekdays that can be constructed = 7C3 = (7.6.5)/(1.2.3) = 35

The correct answer: D

Hope this helps!

-Jay
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[Scott@TargetTestPrep]

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

[spoiler]OA=D[/spoiler]

Source: Magoosh

The number of different three-day combinations of the seven weekdays can be constructed is:

7C3 = (7 x 6 x 5)/(3 x 2) = 7 x 5 = 35

Answer: D