In how many different ways can a committee consisting of 5 men and 6 women be formed from a group of 8 men and 10 women?

Introduction / Context:
This is a combinations problem where we have two separate categories (men and women) and must form a committee with fixed numbers from each category. Such questions test the ability to apply combinations independently to different groups and then combine the results using multiplication.

Given Data / Assumptions:

• Total men available = 8.
• Total women available = 10.
• Required committee size = 11 people.
• Required composition: 5 men and 6 women.
• All individuals are distinct.
• Order of members in the committee is not important.

Concept / Approach:
We must choose 5 men out of 8 and 6 women out of 10. The counts for men and women are independent, so we use the combination formula C(n,r) for each group and then multiply. We do not add because both selections occur together to form one committee. This is the standard approach for forming groups with fixed composition.

Step-by-Step Solution:
Number of ways to choose 5 men from 8 men = C(8,5) = C(8,3) = 56. Number of ways to choose 6 women from 10 women = C(10,6) = C(10,4) = 210. Each valid committee is formed by one choice of men and one choice of women. Total number of committees = C(8,5) * C(10,6) = 56 * 210. Compute 56 * 200 = 11200. Compute 56 * 10 = 560. Total = 11200 + 560 = 11760.

Verification / Alternative check:
We can verify by quickly approximating. C(8,5) is slightly less than 8^3 and C(10,6) is a few hundred, so a product near ten thousand is reasonable. Also, C(8,5) equals C(8,3) and C(10,6) equals C(10,4), which we used to simplify calculations. The multiplication 56 * 210 performed in parts confirms the final figure of 11760. No other compositions such as 4 men and 7 women satisfy the given restriction, so this count is complete.

Why Other Options Are Wrong:

• 11670: This is slightly less than the correct value and usually results from a multiplication slip.
• 12000: A rounded or guessed figure, not produced by the exact combination values.
• 20050: Much too large, likely from misusing permutations or overcounting committees.

Common Pitfalls:
Students sometimes add C(8,5) and C(10,6) instead of multiplying, which is incorrect for building a single combined committee. Others choose the wrong numbers (such as 4 men and 7 women) or use permutations 8P5 and 10P6, which treat order as important. Avoid such mistakes by clearly identifying that committee formation is order independent and by writing each selection as a combination before multiplying.

Final Answer:
The number of different committees that can be formed is 11760.