To solve this problem, we need to determine how many different ways a coach can choose a team of 5 members, consisting of 3 females and 2 males, from a group of 9 males and 10 females.

Step-by-Step Breakdown:

1. Selecting 3 females from 10 females: The number of ways to choose 3 females from 10 is given by the combination formula: $\binom{n}{r} = \frac{n!}{r!(n - r)!}$ where $n$ is the total number of females, and $r$ is the number of females to choose.

   So, the number of ways to select 3 females from 10 is: $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

2. Selecting 2 males from 9 males: Similarly, the number of ways to choose 2 males from 9 males is: $\binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36$

3. Total number of ways to form the team: Since the selections of females and males are independent of each other, we multiply the two results together to find the total number of ways to form the team: $\binom{10}{3} \times \binom{9}{2} = 120 \times 36 = 4320$

Final Answer:

There are 4,320 different ways for the coach to choose the team of five members consisting of three females and two males.

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Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How would the answer change if the coach could select any number of males and females without restriction?
2. What if the team had to consist of 4 males and 1 female?
3. Can you generalize this method to solve for different team sizes?
4. What if the selection was ordered, meaning the positions in the team mattered?
5. How would the approach change if there were fewer or more males and females in the group?

Tip: Combinations are useful when the order of selection doesn't matter, while permutations are used when the order does matter.