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From a group of 7 men and 6 women, five persons are to be selected to

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From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post Updated on: 22 Feb 2019, 03:01
4
13
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A
B
C
D
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  35% (medium)

Question Stats:

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From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?


A) 564
B) 645
C) 735
D) 756
E) 566

Originally posted by azamaka on 26 Sep 2016, 02:38.
Last edited by Bunuel on 22 Feb 2019, 03:01, edited 1 time in total.
Updated.
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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 26 Sep 2016, 03:21
3
azamaka wrote:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?


A) 564
B) 645
C) 735
D) 756
E) 566


Hi
A quick observation..
Ans will have selection of atleast 3 from 7 men, so the answer has to be a multiple of 7..
So ONLY C and D are left..
A high probability to get your answer correct even if you did not know what to do..

Now TWO ways..
1) Now select 3 from 7 and select next 2 from remaining 6 females Or both from males or one each..
And the solution as mentioned above will be
7C5+7C4*6C1+7C3*6C2=21+35*6+35*15=756..

2) Another way total ways - ways where 3 males are not there..
All females 6C5=6..
All but one 6C4*7C1=15*7=105
All but 2..6C3*7C2= 20*21=420
So ways where 3 males are not there 6+105+420=531..

Total 13C5= 13*12*11*10*9/(2*3*4*5)= 13*11*9=1287..
So remaining ways = 1287-531=756..

D
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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 26 Sep 2016, 02:50
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azamaka wrote:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Since at least 3 men must be chosen, we consider all committees which include 3, 4, and 5 men, with 2, 1, and 0 women, respectively. That is, we want to add the number of ways to:

a. Choose 3 from 7 men and 2 from 6 women = (7C3)*(6C2) = 35*15 = 525
b. Choose 4 from 7 men and 1 from 6 women = (7C4)*(6C1) = 35*6 = 210
c. Choose 5 from 7 men and 0 from 6 women = (7C5)*(6C0) = 21*1 = 21

a+b+c= 756. Hence option D

Hope it helps :)
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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 26 Sep 2016, 03:44
1
Total No. of possible combinations possible = (3 Men, 2 Women) + (4 Men, 1 Women) + (5 Men, 0 Women)
= (7C3 * 6C2) + (7C4*6C1) + (7C5)
= 756

Hence, correct answer should be D.
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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 17 Nov 2016, 19:32
Probability (at least 3 men) = prob(3 men, 2 women) + prob(4 men, 1 woman) + prob(5 men) --> (7C3)(6C2)+(7C4)(6C1)+(7C5) = (15)(35)+(6)(35)+21 = 756

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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 29 Nov 2016, 16:40
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1
azamaka wrote:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A) 564
B) 645
C) 735
D) 756
E) 566


We are given a group of 7 men and 6 women and need to determine the number of ways a committee of 5 can be formed with at least 3 men on the committee.

Thus, we have 3 scenarios in which at least 3 men can be selected for the 5-person committee: 3 men and 2 women OR 4 men and 1 woman OR 5 men. Let’s calculate the number of ways to select the committee in each scenario.

Scenario 1: 3 men and 2 women

Number of ways to select 3 men: 7C3 = (7 x 6 x 5)/3! = (7 x 6 x 5)/(3 x 2 x 1) = 35

Number of ways to select 2 women: 6C2 = (6 x 5)/2! = (6 x 5)/(2 x 1) = 15

Thus, the number of ways to select 3 men and 2 women is 35 x 15 = 525.

Scenario 2: 4 men and 1 woman

Number of ways to select 4 men: 7C4 = (7 x 6 x 5 x 4)/4! = (7 x 6 x 5 x 4)/(4 x 3 x 2 x 1) = 35

Number of ways to select 1 woman: 6C1 = 6

Thus, the number of ways to select 4 men and 1 woman is 35 x 6 = 210.

Scenario 3: 5 men

Number of ways to select 5 men: 7C5 = (7 x 6 x 5 x 4 x 3)/5! = (7 x 6 x 5 x 4 x 3)/(5 x 4 x 3 x 2 x 1) = 42/2 = 21

Thus, the number of ways to select a 5-person committee with at least 3 men is:

525 + 210 + 21 = 756

Answer: D
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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 27 Jan 2017, 20:56
Can this be done in a way where we select total and then subtract the "anti-case"
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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 17 Apr 2019, 08:22
azamaka wrote:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?


A) 564
B) 645
C) 735
D) 756
E) 566


7c3*6c2=525
7c4*6c1=210
7c5=21
sum = 756
IMO D
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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 05 May 2019, 06:26
Archit3110 wrote:
azamaka wrote:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?


A) 564
B) 645
C) 735
D) 756
E) 566


7c3*6c2=525
7c4*6c1=210
7c5=21
sum = 756
IMO D


Hi!

Could you please explain what my solution assumes and what exactly it means?

At least 3 men: 7C3 = 35 combinations
2 people from the left ones: 10C2 (10 = 4 men + 6 women are left) = 45 combinations

35 * 45 = 1575

Thank you.
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Re: From a group of 7 men and 6 women, five persons are to be selected to  [#permalink]

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New post 12 May 2019, 04:10
chetan2u wrote:
azamaka wrote:
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?


A) 564
B) 645
C) 735
D) 756
E) 566


Hi
A quick observation..
Ans will have selection of atleast 3 from 7 men, so the answer has to be a multiple of 7..
So ONLY C and D are left..
A high probability to get your answer correct even if you did not know what to do..

Now TWO ways..
1) Now select 3 from 7 and select next 2 from remaining 6 females Or both from males or one each..
And the solution as mentioned above will be
7C5+7C4*6C1+7C3*6C2=21+35*6+35*15=756..

2) Another way total ways - ways where 3 males are not there..
All females 6C5=6..
All but one 6C4*7C1=15*7=105
All but 2..6C3*7C2= 20*21=420
So ways where 3 males are not there 6+105+420=531..

Total 13C5= 13*12*11*10*9/(2*3*4*5)= 13*11*9=1287..
So remaining ways = 1287-531=756..

D


hi,
chetan2u

A quick observation..
Ans will have selection of atleast 3 from 7 men, so the answer has to be a multiple of 7.

Can you please explain above? I could not understand.
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Re: From a group of 7 men and 6 women, five persons are to be selected to   [#permalink] 12 May 2019, 04:10
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