A committee of 2 men and 2 women is to be chosen from a group of 5 men and 6 women. How many different committees are possible?

Choose the men and the women separately, then multiply. The number of ways to pick $2$ of the $5$ men is $\binom{5}{2}=10$, and the number of ways to pick $2$ of the $6$ women is $\binom{6}{2}=15$. The total is $10\cdot15=150$, choice H. Choice G ($30$) multiplies only one of the combinations by something unrelated. Choice F ($20$) is $5\cdot4$, a permutation of the men. Choice J ($330$) is $\binom{11}{4}$, the number of ways to choose $4$ people with no restriction on gender. The committee must have exactly two of each, so the separate counts multiply.