If six persons are to be selected to form a committee from a group of seven women and four men so that at least three women are there on the committee, then the number of ways it can be done, is
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Logic Tip: A faster way using complements: Calculate total unrestricted combinations ($^{11}C_6 = 462$) and subtract the invalid cases. The only invalid cases are those with fewer than 3 women: 2W/4M ($^7C_2 \cdot ^4C_4 = 21 \cdot 1 = 21$). $462 - 21 = 441$.