assignments of positions

Question

involving the assignment of tasks to the men and women of a committee. Assume that you have a committee of 14 members, made up of 6 men and 8 women.
In how many ways can the 3 tasks be assigned so that both men and women are given assignments?

Answer 1

There are two possibilities for assigning the tasks:  1 to a man and 2 to a woman, or 2 to a man and one to a woman.  Let's look at each separately, and then we'll add the results together.

 

1 MAN, 2 WOMAN:  Since there are 6 men, there are ₁C₆ = 6 ways to assign the 1 task to a man.  Since there are 8 women, there are ₂C₈ ways to assign the two tasks to a woman.  That's 8! / (2!6!) = 8*7 / 2 = 28.  Since the assignments to men and women are independent, multiply the number of ways for the man times the number of ways for the women to get the total number of combinations.  That's 6 * 28 = 168.

 

 

2 MEN, 1 WOMAN:  In the same way as above, there are ₂C₆ ways to assign 2 tasks to a man = 6! / (2!4!) = 6*5/2 = 15.  And there are ₁C₈ = 8 ways to assign 1 task to a woman.  That's a total of 15*8 = 120 ways in this case.

 

The total of the two cases is 168 + 120 = 288.