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How many different committees of 2 men and 2 women can be fo
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10 Aug 2020, 10:09
10 Aug 2020, 10:09
Expert Reply
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Question Stats:
78% (00:43) correct
21% (01:03) wrong based on 37 sessions
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How many different committees of 2 men and 2 women can be formed from a group of 12 people, half of whom are men?
A. 225
B. 450
C. 495
D. 900
E. 2,970
A. 225
B. 450
C. 495
D. 900
E. 2,970
Re: How many different committees of 2 men and 2 women can be fo
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11 Aug 2020, 04:31
11 Aug 2020, 04:31
2 of 6 men and 2 of 6 women
6C2 * 6C2 = 225
6C2 * 6C2 = 225
Re: How many different committees of 2 men and 2 women can be fo
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12 Aug 2020, 07:42
12 Aug 2020, 07:42
1
Since we have 12 people available
half of them is 6.
So we have 6 men, and the rest are 6 women
EventA: Choose two men out of the 6 available
EventB: Choose two women out of the 6 available
Number of outcomes in EventA= (6 choose 2)
Number of outcomes in EventB= (6 choose 2)
Why combinations? Because order doesn't matter. A group of John and Jake is the same as a group of Jake and John.
Also the number of outcomes of EventA and EventB are independent of each other. That is, no matter which two guys we pick for EventA there will always be (6 choose 2) options for EventB, and vice-versa.
Therefore we can use Principle of Counting.
Number of total outcomes
= Outcomes in A * Outcomes in B
= (6 choose 2)*(6 choose 2)
= 6!/(4!2!) * 6!/(4!2!)
= 6!6!/(4!4!2!2!)
=(6*5*6*5)/(2*2)
=3*5*3*5
=15*15
= 225
Final Answer: A
half of them is 6.
So we have 6 men, and the rest are 6 women
EventA: Choose two men out of the 6 available
EventB: Choose two women out of the 6 available
Number of outcomes in EventA= (6 choose 2)
Number of outcomes in EventB= (6 choose 2)
Why combinations? Because order doesn't matter. A group of John and Jake is the same as a group of Jake and John.
Also the number of outcomes of EventA and EventB are independent of each other. That is, no matter which two guys we pick for EventA there will always be (6 choose 2) options for EventB, and vice-versa.
Therefore we can use Principle of Counting.
Number of total outcomes
= Outcomes in A * Outcomes in B
= (6 choose 2)*(6 choose 2)
= 6!/(4!2!) * 6!/(4!2!)
= 6!6!/(4!4!2!2!)
=(6*5*6*5)/(2*2)
=3*5*3*5
=15*15
= 225
Final Answer: A
