For this problem, remember that AND means to multiply and OR means to add.
From a group of 6 girls and 7 boys, how many 5-member committees consist of
(a) 3 girls and 2 boys
(6 girls CHOOSE 3) AND (7 boys CHOOSE 2) = (6C3)∙(7C2) = 20∙21 = 420
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(b) 3 boys and 2 girls
(7 boys CHOOSE 3) AND (6 girls CHOOSE 2) = (7C3)∙(6C2) = 35∙15 = 525
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(c) members of the same sex
(6 girls CHOOSE 5) OR (7 boys CHOOSE 5) = (6C5) + (7C5) = 6 + 21 = 26
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(d) more boys than girls.
(6 girls CHOOSE 2) AND (7 boys CHOOSE 3) = (6C2)∙(7C2) = 15∙35 = 525
OR
(6 girls CHOOSE 1) AND (7 boys CHOOSE 4) = (6C1)∙(7C4) = 6∙35 = 210
OR
7 boys CHOOSE 5 = 7C5 = 21
Answer: 525 + 210 + 21 = 756
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One of the boys cannot get along with one of the girls. Find the
number of committees which include this particular boy or girl, but not
both.
Number of committees that leave out the particular girl:
12 people CHOOSE 5 = 12C5 = 792
minus the number that leave out both the particular boy and girl:
11 people CHOOSE 5 = 11C5 = 462
That's 792-462 = 330
OR
Number of committees that leave out the particular boy:
12 people CHOOSE 5 = 12C5 = 792
minus the number that leave out both the particular boy and girl:
11 people CHOOSE 5 = 11C5 = 462
That's 792-462 = 330
Answer: 330 + 330 = 660
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Edwin
