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A committee of 6 is chosen from 8 men and 5 women so as to contain at 02 Apr 2021, 05:15
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71% (02:17) correct 29% (03:02) wrong based on 163 sessions

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A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

A. 635
B. 700
C. 1404
D. 2620
E. 3510
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B
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A committee of 6 is chosen from 8 men and 5 women so as to contain at 02 Apr 2021, 05:23
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Atleast 2M and 3W must be choosen among 6.So that leaves a single position that can be either women or men. So two cases arises:
Case1: 3M and 3W are chosen in 8C3*5C3=(56*10)=560 ways

Case2: 2M and 4W are chosen in 8C2*5C4=(28*5)=140 ways

So a total of 560+140=700 different ways committees can be chosen

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A committee of 6 is chosen from 8 men and 5 women so as to contain at 02 Apr 2021, 05:27
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As the problem says «at least», so, we have two ways: 8C3 × 5C3
And 8C2 ×5C4
==» 560 & 140 =700
Option B

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A committee of 6 is chosen from 8 men and 5 women so as to contain at 02 Apr 2021, 08:59
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Given: A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women.
Asked: How many different committees could be formed?

M-> Men
W-> Women

2M, 4W
8C2*5C4 = 28*5 = 140

3M, 3W
8C3*5C3 = 56*10 = 560

No other cases are feasible

Number of committees that could be formed = 140 + 560 = 700

IMO B
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A committee of 6 is chosen from 8 men and 5 women so as to contain at 05 Apr 2021, 05:15
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Bunuel
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

A. 635
B. 700
C. 1404
D. 2620
E. 3510
Show more

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

So # of committees equals to: \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\).

Answer: B.
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A committee of 6 is chosen from 8 men and 5 women so as to contain at 04 May 2021, 20:54
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Bunuel
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?

A. 635
B. 700
C. 1404
D. 2620
E. 3510
Show more

why cant I first choose 2 men and 3 women, and then choose the 6th person out of the remaining 8 available?: 8C2(5C3)(8)=2240

what is wrong with this reasoning?
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A committee of 6 is chosen from 8 men and 5 women so as to contain at 05 May 2021, 00:45
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Total: 8 + 5 = 13

Men = 8
Women = 5
Committee = 6 people

Condition = at least 2 men and 3 women [2 men and 4 women or 3 men and 3 women]


Case I: 2 + 3 = 5 people and 1 remaining will be a women : \(^8{C_2} * ^5{C_4} = 28 * 5 = 140 \)

Case II: 3 + 3 = 6 people: \(^8{C_3} * ^5{C_3} = 56 * 10 = 560\)

Total: 140 + 560 = 700

Answer B
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