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In how many ways may he choose 2 of the pairs of pants, 3 of the shirt, and 1 of the sweaters to pack ?

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2 Answers

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He can choose 2 pairs of pants in $\binom{4}{2}=\frac{4!}{2!(4-2)!}=6$ ways, 3 of the shirts in $\binom{7}{3}=35$ ways and one of the sweather in 3 ways, complexively he can pack in $6*35*3=630$ ways

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There are C(4,2) = 6 ways to choose 2 pairs of pants.

There are C(7,3) = 35 ways to choose 3 shirts.

There are C(3,1) = 3 ways to choose a sweater.

So in total there are: 6*35*3 = 630 ways to pack.

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