Final answer:
To find the number of ways to form a committee with six members having more women than men, we consider all possible compositions that satisfy this condition (3 women and 2 men, 4 women and 1 man, or all 5 women) and calculate the combinations for each case. We then sum these up to get the total number of ways.
Step-by-step explanation:
Combination Problem of Forming a Committee
The question asks how many ways there are to form a committee of six members from a department containing 10 men and 15 women, given that the committee must have more women than men. To solve this, we consider the possible compositions of the committee that satisfy the condition of having more women than men: the committee could have 3 women and 2 men, or 4 women and 1 man, or all 5 women.
3 women and 2 men: We can choose 3 women out of 15 in C(15,3) ways and 2 men out of 10 in C(10,2) ways. The total combinations for this case would be C(15,3) * C(10,2).
4 women and 1 man: Similarly, we can choose 4 women in C(15,4) ways and 1 man in C(10,1) ways. The total combinations for this case would be C(15,4) * C(10,1).
5 women: Finally, we can select 5 women in C(15,5) ways. Since no men are selected in this case, we simply have the combinations for women, which is C(15,5).
To find the total number of ways to form the committee, we sum up the combinations of all the possible cases:
Total number of ways = C(15,3) * C(10,2) + C(15,4) * C(10,1) + C(15,5)
Where C(n,k) represents the number of combinations of n items taken k at a time. This can be calculated using the formula C(n,k) = n! / (k!(n-k)!), where ! denotes factorial.