Question

there are seven men and six women in line at a salsa dance club.
part 1 of 2
(a) the bouncer is told he can only admit two more men and two more women. how many ways can he choose from those in line?
there are 315 ways he can choose two men and two women.
part: 1 / 2
part 2 of 2
(b) the bouncer is told he can only admit four more people, and at least two must be women. how many ways can he choose from those in line?
there are ways he can choose a group of four with at least two women.

Explanation:

Step1: Calculate total number of people

There are \( 7 + 6=13 \) people in line.

Step2: Calculate the number of ways to choose \( 4 \) people with at least \( 2 \) women

We use the principle of complementary counting. The total number of ways to choose \( 4 \) people out of \( 13 \) is \( C(13,4)=\frac{13!}{4!(13 - 4)!}=\frac{13\times12\times11\times10}{4\times3\times2\times1}=715 \)
The number of ways to choose \( 0 \) women (i.e., \( 4 \) men) is \( C(7,4)=\frac{7!}{4!(7 - 4)!}=\frac{7\times6\times5}{3\times2\times1} = 35 \)
The number of ways to choose \( 1 \) woman (and \( 3 \) men) is \( C(6,1)\times C(7,3)=6\times\frac{7!}{3!(7 - 3)!}=6\times\frac{7\times6\times5}{3\times2\times1}=210 \)

Step3: Calculate the number of ways to choose at least \( 2 \) women

The number of ways to choose at least \( 2 \) women is \( C(13,4)-C(7,4)-C(6,1)\times C(7,3)=715-35 - 210=1050 \)

Answer:

\( 1050 \)