Question 1135741
The problem states that the committee must have a majority of men AND have at least one woman.  That means the committee must be comprised of six men and one woman, five men and two women, or four men and three women.
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Ways a committee with 6 men and 1 woman could be formed:
9C6 * 6C1 = 84 * 6 = 504
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Ways a committee with 5 men and 2 women could be formed:
9C5 * 6C2 = 126 * 15 = 1890
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Ways a committee with 4 men and 3 women could be formed:
9C4 * 6C3 = 126 * 20 = 2520
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Add all these together: 504 + 1890 + 2520 = 4914.
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So, there are 4914 different possible combinations.