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Choose at a conference

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Manager
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Joined: 04 Jun 2010
Posts: 113
Concentration: General Management, Technology
Schools: Chicago (Booth) - Class of 2013
GMAT 1: 670 Q47 V35
GMAT 2: 730 Q49 V41
Followers: 11

Kudos [?]: 103 [0], given: 43

Choose at a conference [#permalink] New post 04 Oct 2010, 21:12
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

82% (02:07) correct 18% (01:19) wrong based on 11 sessions
At a conference, one team is made up of 4 men and 5 women. Four presenters are chosen to present the team's findings in front of the entire conference. How many different groups of presenters can be chosen from the team if a team cannot be composed of men only or women only? (Two groups of presenters are considered different if at least one presenter is different.)

(A) 120
(B) 126
(C) 180
(D) 420
(E) 460


Any quick way to solve this? I only solve it by splitting to 3 different scenarios. Takes way too much time.
[Reveal] Spoiler: OA

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Re: Choose at a conference [#permalink] New post 04 Oct 2010, 22:08
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rafi wrote:
At a conference, one team is made up of 4 men and 5 women. Four presenters are chosen to present the team's findings in front of the entire conference. How many different groups of presenters can be chosen from the team if a team cannot be composed of men only or women only? (Two groups of presenters are considered different if at least one presenter is different.)

(A) 120
(B) 126
(C) 180
(D) 420
(E) 460


Any quick way to solve this? I only solve it by splitting to 3 different scenarios. Takes way too much time.


No of ways = All ways to choose - ways using just men - ways using just women = C(9,4)-C(5,4)-C(4,4) = 126 - 5 - 1 = 120

Answer is (A)
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Manager
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Joined: 19 Jul 2009
Posts: 53
Location: baltimore, md
Schools: kellogg, booth, stern, ann arbor
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Kudos [?]: 19 [0], given: 3

Re: Choose at a conference [#permalink] New post 05 Oct 2010, 08:14
i got A. but i think i did it the long way.

if we want at least 1 sex per group, then we can have
3 women, 1 man
3 men, 1 woman
2 women, 2 men

3 women, 1 man = 5C3 * 4C1 = 40
3 men, 1 woman = 4C3 * 5C1 = 20
2 women, 2 men = 5C2 * 4C2 = 60

40 + 20 + 60 = 120 total
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Re: Choose at a conference   [#permalink] 05 Oct 2010, 08:14
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