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In how many ways can a committee consisting of 2 men and at least 2 women be chosen from 4 men and 3 women?
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two men and at least 2 women means a) two men and two women b) two men and three women we can compute each
2 men, 2 women \[\binom{4}{2}\times \binom{3}{2}\]i.e "four choose two times three choose two
2 men, 3 women \[\binom{4}{2}\times \binom{3}{3}\] which is just \(\binom{4}{2}\) since three choose 3 is 1
you know how to compute these?
no
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have you seen "n choose k" sometimes written as \[_nC_k\] or \[\binom{n}{k}\]?
yes
Could you pls help me to get answers for those questions
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