A student club consists of 8 guys and 12 girls. A committee of 8 is to be selected. In how many ways can this be done if
the committee must have the same number of guys as girls
Order doesn't matter, so we use combinations:
Choose the 4 guys 8C4 ways and the 4 girls 12C4 ways.
Answer: (8C4)(12C4) = (70)(495) = 34560
the committee must have at least 6 guys on it
Case 1: 6 guys and 2 girls.
Choose the 6 guys 8C6 ways and the 2 girls 12C4 ways.
Total for case 1: (8C6)(12C2) = (28)(66) = 1848.
Case 2: 7 guys and 1 girl.
Choose the 7 guys 8C7 ways and the 1 girls 12C1 ways.
Total for case 3: (8C7)(12C1) = (8)(12) = 96.
Case 3: 8 guys and no girls.
Choose the 8 guys 8C8 ways.
Total for case 3: (8C8)(12C2) = 1.
Total for all three cases = 1848+96+1 = 1945 ways.
three officers (president, vice president, secretary/treasurer) are selected from ALL the members.
Here order matters, so we use permutations: 20P3 = 6840 ways.
Or you can think of it this way:
Choose the president any of 20 ways.
Choose the vice president any of the remaining 19 ways.
Choose the vice secretary/treasurer any of the remaining 18 ways.
Answer: 20*19*18 = 6840
Edwin
