In how many ways can the volunteers be assigned

[wind]

The social convenor has 12 volunteers to work at a school dance. Each dance requires 2 volunteers at the door, 4 volunteers on the floor, and 6 floaters. Joe and Jim have not volunteered before, so the social covenor does not want to assign them to work together. In how many ways can the volunteers be assigned?

 

[stapel]

I think you could do this by considering cases.

. . . . .Joe at the door
. . . . .Jim on the floor

This leaves ten people who can be put anywhere:

. . . . .door: ₁₀C₁

This leaves nine people who can be put in either of the other locations:

. . . . .floor: ₉C₃

This leaves six as floaters:

. . . . .floating: ₆C₆

Multiply to get the numbers of ways to assign people, assuming Jim and Joe are assigned as decided above.

Now figure out the other cases in the same way. Add up the numbers for each case to find the total number of options.

There may be (and probably is) a neater way of doing this, so, if somebody else replies with another method, go with that instead.

Eliz.

 

[pka]

The following equals the number of ways to assign both men to the same group:
\(\displaystyle \underbrace {\left( {\begin{array}{c}
{10} \\
4 \\
\end{array}} \right)}_{{\rm{door}}} + \underbrace {\left( {\begin{array}{c}
{10} \\
2 \\
\end{array}} \right)\left( {\begin{array}{c}
8 \\
2 \\
\end{array}} \right)}_{{\rm{floor}}} + \underbrace {\left( {\begin{array}{c}
{10} \\
2 \\
\end{array}} \right)\left( {\begin{array}{c}
8 \\
4 \\
\end{array}} \right)}_{{\rm{float}}}\).

Those are the cases we do not want.
Remove them from the total: \(\displaystyle \left( {\begin{array}{c}
{12} \\
2 \\
\end{array}} \right)\left( {\begin{array}{c}
{10} \\
4 \\
\end{array}} \right)\).