A committee is made up of 12 students. Five of them are seniors and the rest are juniors.
2026-03-31 23:32:52.1774999972
in how many ways can a subcommittee of five students be chosen if exactly three of them must be seniors and the rest juniors?
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If there are $5$ seniors then there must be $7$ juniors. We can pick three seniors out of five in $^5 C_{3} = \binom{5}{3} = 10$ ways. Similarly, we can pick $5-3=2$ juniors from seven juniors in $^7 C_{2} = \binom{7}{2} = 21$ ways. Using multiplication, we have a total of $$21\times 10= 210$$ ways. Hope it helps.