A group consists of 4 men, 6 women and 5 children.\nIn how many different ways can exactly 2 men, 3 women and 1 child be selected from this group?

Introduction / Context:
This question is a straight application of combinations with several independent groups. We have to form a small team with fixed numbers chosen from men, women and children. The main idea is that order does not matter within the team, so combinations rather than permutations are used, and choices from each subgroup are independent of each other.

Given Data / Assumptions:

• Total men in the group = 4.
• Total women in the group = 6.
• Total children in the group = 5.
• Required composition of the team: 2 men, 3 women and 1 child.
• Each person is distinct.
• Order of people inside the selected team does not matter.

Concept / Approach:
When we choose separate categories independently, we use combinations within each category and then multiply the results. So we calculate how many ways to choose 2 of the 4 men, 3 of the 6 women and 1 of the 5 children. The rule used is the combination formula C(n,r) = n! / (r! * (n - r)!). Finally, we multiply the three combination values because any valid team is obtained by combining one choice of men, one choice of women and one choice of children.

Step-by-Step Solution:
Number of ways to choose 2 men from 4 men = C(4,2) = 4! / (2! * 2!) = 6. Number of ways to choose 3 women from 6 women = C(6,3) = 6! / (3! * 3!) = 20. Number of ways to choose 1 child from 5 children = C(5,1) = 5. Total ways to form the required team = C(4,2) * C(6,3) * C(5,1) = 6 * 20 * 5. First compute 6 * 20 = 120. Then 120 * 5 = 600.

Verification / Alternative check:
We can double check by quickly reasoning about size: the largest combination used is C(6,3) = 20, and the product 6 * 20 * 5 stays comfortably below 1000, which is reasonable for such a small group. There is no other way to satisfy the exact counts 2, 3 and 1 than by choosing independently from each subgroup, so there is no overcount or undercount. This confirms that 600 is the correct number of possible teams.

Why Other Options Are Wrong:

• 480: This often comes from using C(4,2) * C(6,2) * C(5,1), which incorrectly chooses only 2 women instead of 3.
• 720: This is commonly from multiplying an extra factor such as 6! without justification, mixing permutations with combinations.
• 540: This is simply an arithmetic error, for example miscalculating 6 * 20 * 5.

Common Pitfalls:
Typical mistakes include using permutations instead of combinations, choosing wrong counts from each subgroup, or adding rather than multiplying the individual combination values. Students sometimes also forget that one child must be chosen and wrongly treat children as optional. Writing each stage clearly as C(4,2), C(6,3) and C(5,1), and then multiplying step by step, helps avoid these errors. Always check that the final number matches the size and structure of the selection problem.

Final Answer:
The required number of different ways to select 2 men, 3 women and 1 child is 600.