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serious [3.7K]
2 years ago
15

A committee consists of 8 men and 11 women. In how many ways can a subcommittee of 3 men and 5 women be chosen?

Mathematics
1 answer:
PIT_PIT [208]2 years ago
8 0

Answer:

25872 ways

Step-by-step explanation:

We're choosing 5 women from a group of 11 and 3 men from a group of 8. We don't care about what order they are picked and so we'll use the combination formula, which is:

n!/(k!)(n-k)! with n as population and k as picks.

We'll multiply the results together. (8! / (3!)(8-3)!) * (11! / (5!)(11-5)!)

That equals: (8! / (3!)(5!) ) * (11! / (5!)(6!)) = 40320/(6x120) * 39916800/ (120x720)

56 * 462 = 25872

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3 years ago
What are the possible rational zeros for f(x)=3x^4+x^3-13x^2-2x+9?
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Answer:

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Step-by-step explanation:

The given expression is f(x)=3x^{4}+x^{3}-13x^{2}-2x+9

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