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# Four representatives are to be randomly chosen from a committee of six

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Math Expert
Joined: 02 Sep 2009
Posts: 42246

Kudos [?]: 132652 [0], given: 12331

Four representatives are to be randomly chosen from a committee of six [#permalink]

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28 Sep 2016, 03:00
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Question Stats:

78% (01:37) correct 22% (01:55) wrong based on 117 sessions

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Four representatives are to be randomly chosen from a committee of six men and six women. How many different 4-representative combinations contain at least one woman?

A. 400
B. 415
C. 455
D. 480
E. 495
[Reveal] Spoiler: OA

_________________

Kudos [?]: 132652 [0], given: 12331

Intern
Joined: 29 Aug 2016
Posts: 4

Kudos [?]: [0], given: 108

Re: Four representatives are to be randomly chosen from a committee of six [#permalink]

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28 Sep 2016, 03:14
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Bunuel wrote:
Four representatives are to be randomly chosen from a committee of six men and six women. How many different 4-representative combinations contain at least one woman?

A. 400
B. 415
C. 455
D. 480
E. 495

Should be D

12*11*10*9/4*3*2*1=11*5*9=495

Forbidden options
6*5*4*3/4*3*2*1=5*3=15

495-15=480

A good one tough question

Kudos [?]: [0], given: 108

Intern
Joined: 21 Dec 2014
Posts: 7

Kudos [?]: 14 [2], given: 7

Re: Four representatives are to be randomly chosen from a committee of six [#permalink]

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28 Sep 2016, 06:08
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Bunuel wrote:
Four representatives are to be randomly chosen from a committee of six men and six women. How many different 4-representative combinations contain at least one woman?

A. 400
B. 415
C. 455
D. 480
E. 495

Total Ways in which any 4 representatives can be selected (T) = $$^{12}C_4$$ (Total Members = 12 (6 Men and 6 Women) )
Ways in which all men are selected (M) = $$^{6}C_4$$

Ways in which at least one woman representative is selected (W) = T - M
$$W = {^{12}C_4} - {^{6}C_4} = {495 - 15} = 480$$ Answer : D

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Kudos [?]: 14 [2], given: 7

Intern
Joined: 06 Jul 2017
Posts: 6

Kudos [?]: 0 [0], given: 22

Re: Four representatives are to be randomly chosen from a committee of six [#permalink]

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24 Sep 2017, 05:36
Hey Guys,

How come that I can not just add all single possibilities of how the committee is structured e.g. (All Females) + (3 Females 1 Male) + (2 Females 2 Males) + (1 Female 3 Males):

n= 6C4 + (6C3 *6C1) + (6C2 * 6C2) + (6C1 * 6C3) ?

Kudos [?]: 0 [0], given: 22

Manager
Joined: 17 May 2015
Posts: 208

Kudos [?]: 239 [1], given: 73

Re: Four representatives are to be randomly chosen from a committee of six [#permalink]

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24 Sep 2017, 06:00
1
KUDOS
oli29 wrote:
Hey Guys,

How come that I can not just add all single possibilities of how the committee is structured e.g. (All Females) + (3 Females 1 Male) + (2 Females 2 Males) + (1 Female 3 Males):

n= 6C4 + (6C3 *6C1) + (6C2 * 6C2) + (6C1 * 6C3) ?

Hi oli29,

Your method is perfectly fine. The above expression would give you the same result, but as you notice that the calculation becomes tedious.

There is a small trick to solve such problems: Use Complementary Method.

Total number of ways to choose four representative = (4 Females, 0 Males) + (3 Females, 1 Male) + (2 Females, 2 Males) + (1 Female, 3 Males) + (0 Females, 4 Males).

Please note that required event = Total - (0 Females, 4 Males).

Please refer the post by TheOutlawTorn.

Hope this helps.

Kudos [?]: 239 [1], given: 73

Intern
Joined: 06 Jul 2017
Posts: 6

Kudos [?]: 0 [0], given: 22

Re: Four representatives are to be randomly chosen from a committee of six [#permalink]

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24 Sep 2017, 06:13
ganand wrote:
oli29 wrote:
Hey Guys,

How come that I can not just add all single possibilities of how the committee is structured e.g. (All Females) + (3 Females 1 Male) + (2 Females 2 Males) + (1 Female 3 Males):

n= 6C4 + (6C3 *6C1) + (6C2 * 6C2) + (6C1 * 6C3) ?

Hi oli29,

Your method is perfectly fine. The above expression would give you the same result, but as you notice that the calculation becomes tedious.

There is a small trick to solve such problems: Use Complementary Method.

Total number of ways to choose four representative = (4 Females, 0 Males) + (3 Females, 1 Male) + (2 Females, 2 Males) + (1 Female, 3 Males) + (0 Females, 4 Males).

Please note that required event = Total - (0 Females, 4 Males).

Please refer the post by TheOutlawTorn.

Hope this helps.

Thanks allot, I kept making a stupid calc mistake. Yes the above is certainly quicker and I will try to do the questions now like this. However I just wanted to confirm the underlying concept.

Kudos [?]: 0 [0], given: 22

Target Test Prep Representative
Affiliations: Target Test Prep
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Kudos [?]: 902 [0], given: 5

Re: Four representatives are to be randomly chosen from a committee of six [#permalink]

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26 Sep 2017, 16:40
Bunuel wrote:
Four representatives are to be randomly chosen from a committee of six men and six women. How many different 4-representative combinations contain at least one woman?

A. 400
B. 415
C. 455
D. 480
E. 495

We can determine the number of ways to select 4-representative combinations containing at least one woman. Let’s use the formula:

Total number of ways to select a group of 4 - number of ways to form the group with no women

Total number of ways to select a group of 4 is 12C4 = 12!/[4!(12-4)!] = 12!/(4! X 8!) = (12 x 11 x 10 x 9)/4! = (12 x 11 x 10 x 9)/(4 x 3 x 2 x 1) = 11 x 5 x 9 = 495.

Number of ways to form the group with no women = 6C4 = 6!/[4!(6-4)!] = 6!/(4! x 2!) = (6 x 5 x 4 x 3)/4! = (6 x 5 x 4 x 3)/(4 x 3 x 2 x 1) = 5 x 3 = 15.

Thus, the total number of ways to select the group with at least one woman is 495 - 15 = 480.

_________________

Jeffery Miller

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Kudos [?]: 902 [0], given: 5

Re: Four representatives are to be randomly chosen from a committee of six   [#permalink] 26 Sep 2017, 16:40
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