A man has 9 friends (4 boys and 5 girls). In how many ways can he invite a group that contains exactly 3 girls (with any number of boys, including possibly none)?

Introduction / Context:
The invitation must include exactly 3 girls, while the number of boys is unconstrained. We count subsets: first fix the girls, then freely choose any subset of boys.

Given Data / Assumptions:

• Girls = 5, Boys = 4.
• Pick exactly 3 girls.
• Pick any subset of the 4 boys (0 through 4 allowed).

Concept / Approach:
Multiply ways to choose girls by ways to (independently) choose boys. The number of subsets of 4 boys is 2^4. The number of 3-girl choices is C(5,3).

Step-by-Step Solution:
C(5,3) = 10.Subsets of 4 boys = 2^4 = 16.Total invitation groups = 10 * 16 = 160.

Verification / Alternative check:
Enumerating boy counts (0..4) gives Σ C(5,3)*C(4,k) = C(5,3) * 2^4 by the binomial theorem.

Why Other Options Are Wrong:
320 double counts by also forcing boys; 200 and 80 arise from partial sums or mixing constraints.

Common Pitfalls:
Assuming a fixed total size or forgetting to include the empty boy subset.

Final Answer:
160