We are given a group of 4 girls and 7 boys, and we need to select a team of 5 members such that no girls are included. In other words, the team must consist of only boys.

Since there are 7 boys in total, and the team must have 5 members, we need to select 5 boys from the 7 boys. The number of ways to choose 5 boys out of 7 is given by the combination formula:

$\text{Number of ways} = \binom{7}{5}$

The formula for combinations is:

$\binom{n}{r} = \frac{n!}{r!(n - r)!}$

Now, apply it to our case:

$\binom{7}{5} = \frac{7!}{5!(7 - 5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$

Thus, the number of ways to select a team of 5 members with no girls is 21.

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Related Questions:

1. How many ways can a team be selected if there must be at least 1 girl?
2. How many ways can a team be selected if there are exactly 2 girls?
3. What if the team must include exactly 1 boy and 4 girls?
4. How many ways can we select a team of 5 members if the team must include at least 3 boys?
5. How many different teams can be selected from the entire group of 11 members?

Tip:

For problems involving selection with restrictions, breaking the group into smaller subgroups (e.g., girls and boys) and handling each separately simplifies the calculation.