Mathematics 2010 JAMB Past Questions - Question 49: In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women?

Correct Answer

B

Explanation

5C₂ * 6C₃ = 5 * 4 /2*1 X 6* 5 *4/3*2 *1 = 10 20 = 200To determine the number of ways a committee of 2 women and 3 men can be chosen from a group of 6 men and 5 women, we can use the concept of combinations.The number of ways to choose 2 women from a group of 5 women is given by the combination formula:C(5, 2) = 5! / (2! * (5 - 2)!) = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10Similarly, the number of ways to choose 3 men from a group of 6 men is given by the combination formula:C(6, 3) = 6! / (3! * (6 - 3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20To find the total number of ways to choose 2 women and 3 men, we multiply the number of ways to choose women by the number of ways to choose men:Total number of ways = 10 * 20 = 200Therefore, there are 200 ways to choose a committee of 2 women and 3 men from a group of 6 men and 5 women.