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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 26 Sep 2016, 02:38
3
3
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

77% (02:32) correct 23% (03:00) wrong based on 148 sessions

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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
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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 26 Sep 2016, 02:50
3
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?

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, 03:21
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


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, 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
1
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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