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The answer is 10 over 3 (you write 10 over 3, without a fraction line in between, and with parentheses around the entire expression). This is calculated as (10 x 9 x 8) / (1 x 2 x 3).

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Q: In how many ways can a committee of 3 be chosen from 10 students?
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The first member chosen can be any one of 1,514 students.The second member chosen can be any one of the remaining 1,513 students.The third member chosen can be any one of the remaining 1,512 students.So there are (1,514 x 1,513 x 1,512) ways to choose three students.But for every group of three, there are (3 x 2 x 1) = 6 different orders in which the same 3 can be chosen.So the number of `distinct, unique committees of 3 students is(1514 x 1513 x 1512) / 6 = 577,251,864


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In how many ways can a committee of three men and four women be formed from a group of 10 men and 10 women?

I'm going with 25,200 3 men out of 10 may be chosen in 10C3 ways = 10 ! / 3! 7 ! = 120 ways. 4 women may be chosen out of 10 in 10C4 = 10 ! / 4! 6! ways = 210 ways. Therefore, a committee with 3 men and 4 women can be formed in 120 x 210 = 25,200 ways.