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# 6 male and 8 female employees. A group has to be formed

Author Message
TAGS:
CEO
Joined: 21 Jan 2007
Posts: 2766
Location: New York City
Followers: 9

Kudos [?]: 333 [0], given: 4

6 male and 8 female employees. A group has to be formed [#permalink]  22 Oct 2007, 07:16
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
6 male and 8 female employees. A group has to be formed including 2 men and 1 woman, How many possible variants?

60
80
120
150
180
Manager
Joined: 18 Jun 2007
Posts: 55
Followers: 1

Kudos [?]: 1 [0], given: 0

I'll give it a shot:
8 females, pick only 1. So i did: 8!/(1!7!) = 8
6 males, pick 2. 6!/(2!4!)= (6*5)/2 = 15

8*15 = 120
CEO
Joined: 21 Jan 2007
Posts: 2766
Location: New York City
Followers: 9

Kudos [?]: 333 [0], given: 4

ben928 wrote:
I'll give it a shot:
8 females, pick only 1. So i did: 8!/(1!7!) = 8
6 males, pick 2. 6!/(2!4!)= (6*5)/2 = 15

8*15 = 120

OA is C. 120

6c2 = 15
8c1= 8

15*8= 120
CEO
Joined: 29 Mar 2007
Posts: 2591
Followers: 16

Kudos [?]: 226 [0], given: 0

Re: PS Easy Combo [#permalink]  22 Oct 2007, 21:13
bmwhype2 wrote:
6 male and 8 female employees. A group has to be formed including 2 men and 1 woman, How many possible variants?

60
80
120
150
180

6!/2!4! = number of possibilities for 2 men.

8!/7!1! = number of possibilities for 1 woman.

15*8=120.

C.
Re: PS Easy Combo   [#permalink] 22 Oct 2007, 21:13
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