Combination Question 2?

Question

12 volunteers are to work at a school dance. Each dance requires 2 volunteers at the door, 4 volunteers on the floor, and 6 floaters. If 2 of these volunteers can't work together, then how many ways can the volunteers be assigned?

Thanks

Answer 1

Since 2 + 4 + 6 = 12, it's impossible to have a dance withouting counting on the 2 volunteers that can't work together. Call them a and b. We have C(3, 2) (3 choose 2) = 3 possibilities to assign positions to a and b: door and floor (DF), door and floater (DFL) and floor and floater (FFL)

DF - once we've set a and b, we have to choose 1 volunteer for the door, 3 for the floor and 3 for floaters out of the remaining 10 volunteers. We have C(10,1) = 10 ways to assign one to the door. Choosen this volunteerr for the door, we have, by a similar reasoning, C(9, 3) ways to choose the 3 volunteers for the floor. Once they are choosen, then, since neither a nor b is a floater, we have C(6, 6) = 1 ways to select the floaters. And since we can interchange a nd b, we have
2 * 10 * C9,3) = 20C(9,3) for the volunteers to be assigned in the DF scheme.

DFL - By a similar reasoning, we have 2 * C(10, 1) * C(9, 5) * C4,4) = 20C(9,5) ways.

FFL - similarly, we have 2 * C(10, 3) * C(7, 5) * C(2,2) = 2 * C(10, 3) * C(7, 5) ways.

The sets DF, DFL and FFL are pairwise disjoint and their union covers all the possibilities. So, we have 20 C(9,3) + 20C(9,5) + 2 * C(10, 3) * C(7, 5) ways for the volunteers to be assigned.