Answer:
a) 250
b) 27405
c) 10098
Step-by-step explanation:
a) 12 * 7 * 6 * 5 = 250 committees
b) [tex]\frac{30!}{4! * (30 - 4)!} = \frac{30!}{4! * (26)!} = 27405[/tex] committees
c) Number of ways to select seniors -> [tex]\frac{12!}{2! * 10!} = 66[/tex]
Number of ways to select remaining members -> [tex]\frac{18!}{2! * 16!} = 153[/tex]
66 * 153 = 10098
A - If the committee must contain one person from each class, there would be 2,520 possible committees.
B- If the committee could contain any mixture of the classes, there would be 657,720 possible committees.
C- If the committee must contain exactly two seniors, there would be 40,392 possible committees.
Given that a group of twelve seniors, seven juniors, six sophomores, and five freshmen must select a committee of four, to determine how many committees are possible if the committee must contain (A) one person from each class, (B) any mixture of the classes, and (C) exactly two seniors, the following calculations must be performed:
Therefore, if the committee must contain one person from each class, 2,520 committees are possible; if it can contain any mixture of the classes, 657,720 committees are possible; and, if it must contain exactly two seniors, 40,392 committees are possible.
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