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# Choose at a conference

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Manager
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: 14

Kudos [?]: 225 [0], given: 43

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04 Oct 2010, 22:12
00:00

Difficulty:

25% (medium)

Question Stats:

83% (02:19) correct 17% (01:19) wrong based on 12 sessions

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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]

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04 Oct 2010, 23:08
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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

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Manager
Joined: 19 Jul 2009
Posts: 52
Location: baltimore, md
Schools: kellogg, booth, stern, ann arbor
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Re: Choose at a conference [#permalink]

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05 Oct 2010, 09:14
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, 09:14
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# Choose at a conference

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