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Country A and Country B will f
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10 Jan 2022, 03:14
10 Jan 2022, 03:14
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Question Stats:
60% (01:55) correct
40% (02:38) wrong based on 10 sessions
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Country A and Country B will form a joint committee on economic policy. The committee is to have exactly 6 members. 5 candidates for the committee come from Country A and 6 from Country B. If at least 3 members of the committee must come from Country A, how many distinct committees are possible?
11
30
174
200
281
11
30
174
200
281
Country A and Country B will f
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22 Jan 2022, 18:14
22 Jan 2022, 18:14
1
In this problem, the order of the 6 committee members does not matter, so we can use combinations to solve it.
If 5 candidates for the committee come from Country A and at least 3 members of the committee must come from Country A, then we use the following combinations:
5C3
5C4
5C5
If 6 candidates come from Country B, then we would multiply the Country B combinations with Country A combinations, making sure that each set of combinations adds up to 6 committee members:
5C3(6C3)
5C4(6C2)
5C5(6C1)
Multiply the combination products and add up the distinct committees:
5C3(6C3) = (5*4*3)/(3*2*1)[(6*5*4)/(3*2*1)]= (10)(20) = 200
5C4(6C2) = (5*4*3*2)/(4*3*2*1)[(6*5)/(2*1)] = (5)(15) = 75
5C5(6C1) = (5*4*3*2*1)/(5*4*3*2*1)[(6/1)] = 1(6) = 6
Total Distinct Committees:
200 + 75 + 6 = 281
If 5 candidates for the committee come from Country A and at least 3 members of the committee must come from Country A, then we use the following combinations:
5C3
5C4
5C5
If 6 candidates come from Country B, then we would multiply the Country B combinations with Country A combinations, making sure that each set of combinations adds up to 6 committee members:
5C3(6C3)
5C4(6C2)
5C5(6C1)
Multiply the combination products and add up the distinct committees:
5C3(6C3) = (5*4*
5C4(6C2) = (5*
5C5(6C1) = (
Total Distinct Committees:
200 + 75 + 6 = 281
Re: Country A and Country B will f
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29 Apr 2024, 16:57
29 Apr 2024, 16:57
superpower101 wrote:
In this problem, the order of the 6 committee members does not matter, so we can use combinations to solve it.
If 5 candidates for the committee come from Country A and at least 3 members of the committee must come from Country A, then we use the following combinations:
5C3
5C4
5C5
If 6 candidates come from Country B, then we would multiply the Country B combinations with Country A combinations, making sure that each set of combinations adds up to 6 committee members:
5C3(6C3)
5C4(6C2)
5C5(6C1)
Multiply the combination products and add up the distinct committees:
5C3(6C3) = (5*4*3)/(3*2*1)[(6*5*4)/(3*2*1)]= (10)(20) = 200
5C4(6C2) = (5*4*3*2)/(4*3*2*1)[(6*5)/(2*1)] = (5)(15) = 75
5C5(6C1) = (5*4*3*2*1)/(5*4*3*2*1)[(6/1)] = 1(6) = 6
Total Distinct Committees:
200 + 75 + 6 = 281
If 5 candidates for the committee come from Country A and at least 3 members of the committee must come from Country A, then we use the following combinations:
5C3
5C4
5C5
If 6 candidates come from Country B, then we would multiply the Country B combinations with Country A combinations, making sure that each set of combinations adds up to 6 committee members:
5C3(6C3)
5C4(6C2)
5C5(6C1)
Multiply the combination products and add up the distinct committees:
5C3(6C3) = (5*4*
5C4(6C2) = (5*
5C5(6C1) = (
Total Distinct Committees:
200 + 75 + 6 = 281
Additionally the first calculation (200) immediately tells you that the answer is at least 200. This immediately eliminates all answer choices except 281
