Correct Answer: (C)

1. Identify Possible Cases: Since the committee must have 5 members with at least 2 men and 2 women, the possible compositions are: Case I: 2 Men and 3 Women; Case II: 3 Men and 2 Women.

2. Calculate Case I (2M, 3W): The number of ways to select 2 men from 6 and 3 women from 8 is $\binom{6}{2} \times \binom{8}{3} = 15 \times 56 = 840$.

3. Calculate Case II (3M, 2W): The number of ways to select 3 men from 6 and 2 women from 8 is $\binom{6}{3} \times \binom{8}{2} = 20 \times 28 = 560$.

4. Total Combinations: Adding the two mutually exclusive cases gives $840 + 560 = 1400$. Let us re-verify the selection calculation. $\binom{6}{2}=15$, $\binom{8}{3}=56$, $\binom{6}{3}=20$, $\binom{8}{2}=28$. $15 \times 56 = 840$, $20 \times 28 = 560$. $840+560=1400$. Wait, checking the options and calculation: $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. $\binom{8}{2} = \frac{8 \times 7}{2} = 28$. $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$. $\binom{6}{2} = 15$. If the total is 1400 and not in options, let's re-read: "committee of 5". If the option is 1470, perhaps the pool or committee size is different. However, based on these exact constraints, 1400 is the correct count.

Test Prep Tip: In combination problems with "at least" constraints, always list all mutually exclusive cases clearly before applying the addition principle. Ensure you do not over-count by assuming a specific order of selection.