HW Score: $$ 73.33 \%, 11 $$ of 15 points From the 9 male and 8 female sales representatives for an insurance company, a team of 3 men and 2 women will be selected to attend a national conference on insurance fraud. In how many ways can the team of 5 be selected?

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1. First, select $$3$$ men out of $$9$$. The number of ways to do this is given by:
   $$
   \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = 84
   $$

2. Next, select $$2$$ women out of $$8$$. The number of ways to do this is:
   $$
   \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} = 28
   $$

3. Since these selections are independent, multiply the two results to get the total number of teams:
   $$
   84 \times 28 = 2352
   $$

Thus, the number of ways to select the team is $$2352$$.
There are 2,352 ways to select the team.

HW Score: of 15 points
From the 9 male and 8 female sales representatives for an insurance company, a team of 3 men
and 2 women will be selected to attend a national conference on insurance fraud.
In how many ways can the team of 5 be selected?

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HW Score: of 15 points From the 9 male and 8 female sales representatives for an insurance company, a team of 3 men and 2 women will be selected to attend a national conference on insurance fraud. In how many ways can the team of 5 be selected?