How many ways to organize 12 volunteers with restrictions

Question

You have 12 volunteers for a school dance. You must have 2 volunteers at the door, 4 volunteers on the floor, and 6 floaters. Joe and Jim have never volunteered before, so they cannot be assigned to work together. In how many ways can the volunteers be assigned, with said restrictions?

This is how I've found the total without restrictions, from here I'm unsure of how to continue.

$$\binom{12}{2}\cdot \binom{12}{4}\cdot \binom{12}{6}$$

I'm guessing you would use an indirect method, so treat them as one person so that there are now 11 volunteers, then subtract this from the total number without restrictions. This proves incorrect, as the correct answer in my textbook is 9240 possible ways.

Answer 1

Your idea of subtracting the number of bad assignments from the total number of assignments is sound. However, you have not correctly counted the number of possible assignments and treating Joe and Jim as a single person does not work.

To find the number of possible assignments, assign two of the twelve people to work the door, four of the remaining ten people to work the floor, and all six of the remaining people to be floaters.

This can be done in $$\binom{12}{2}\binom{10}{4}\binom{6}{6}$$ ways.

There are three impermissible assignments:

1. Joe and Jim are both assigned to the door.
2. Joe and Jim are both assigned to the floor.
3. Joe and Jim are both assigned to be floaters.

Joe and Jim are both assigned to the door: We must assign four of the ten remaining people to the floor and all six of the others to be floaters.

This can be done in $$\binom{10}{4}\binom{6}{6}$$ ways.

Joe and Jim are both assigned to the floor: Since four people are assigned to the floor, we must assign two of the remaining ten people to the floor, two of the remaining eight people to the door, and all six of the remaining people to be floaters.

This can be done in $$\binom{10}{2}\binom{8}{2}\binom{6}{6}$$ ways.

Joe and Jim are both assigned to be floaters: Since six people are assigned to be floaters, we must assign four of the remaining ten people to be floaters, four of the remaining six people to work the floor, and both of the remaining two people to the door.

This can be done in $$\binom{10}{4}\binom{6}{4}\binom{2}{2}4$$ ways.

To complete the problem, subtract the three prohibited cases from the total number of possible assignments.