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Intern
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Joined: 06 May 2012
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Perms + Combination [#permalink] New post 20 May 2012, 12:19
A certain club has 5 american, 4 canadian and 3 french members. How many possible basketball teams can be formed if each team must be made up of 2 american, 2 canadian and 1 french member?
a. 4
b. 19
c. 60
d. 180
e. 720

Please help!! I will appreciate your support!
Thks!!
Director
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GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740
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Re: Perms + Combination [#permalink] New post 20 May 2012, 18:01
Select 2 Americans from 5 Americans in 5C2 = 10 ways
Select 2 Canadians from 4 Canadians in 4C2 = 6 ways
Select 1 French from 3 French in 3C1 = 3 ways

Total number of teams = 10 x 6 x 3 = 180

Option (D)
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Manager
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Joined: 21 May 2012
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Location: United States (CA)
Concentration: Marketing, Operations
Schools: Anderson FEMBA '16, Marshall PM '16
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Re: Perms + Combination [#permalink] New post 21 May 2012, 09:23
Looks like this if you break it down by factorials:

(5!)/(2!3!) * (4!)/(2!2!) * (3!)/(1!2!)

120/12 * 24/4 * 6/2

10 * 6 * 3 = 180
Re: Perms + Combination   [#permalink] 21 May 2012, 09:23
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