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

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

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New post 28 Sep 2016, 03:00
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

81% (02:09) correct 19% (02:51) wrong based on 192 sessions

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

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New post 28 Sep 2016, 06:08
5
3
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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Re: Four representatives are to be randomly chosen from a committee of six  [#permalink]

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New post 28 Sep 2016, 03:14
1
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
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Re: Four representatives are to be randomly chosen from a committee of six  [#permalink]

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

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New post 24 Sep 2017, 06:00
1
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.
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Re: Four representatives are to be randomly chosen from a committee of six  [#permalink]

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New post 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.

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

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New post 26 Sep 2017, 16:40
2
1
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.

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

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New post 04 Jun 2019, 20:18
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


At Least one woman = All possible committees - No women committees(Or all men committee)
=> 12C4 - 6C4
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Re: Four representatives are to be randomly chosen from a committee of six   [#permalink] 04 Jun 2019, 20:18
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