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 convenor does not want to assign them to work together. In how man ways can the volunteers be assigned?
I'm totally lost....my multiple attempts have all been significantly higher :s
Any help would be appreciated! Thanks
I'm totally lost....my multiple attempts have all been significantly higher :s
Any help would be appreciated! Thanks
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First let's ignore Joe and Jim, and count all the ways to assign people. There are 12C2 ways to choose the volunteers at the door; after that, there are 10C4 ways to choose the volunteers on the floor, and then the remaining volunteers are floaters. Therefore the total number of arrangements is
12C2 * 10C4 = 66 * 210 = 13860.
Now count the ways to put Joe and Jim together at the door. If you do that, then there are 10 volunteers remaining, so there are 10C4 ways to assign the floor volunteers, and then the rest are floaters. Therefore there are 10C4 = 210 arrangements with Joe and Jim together at the door.
Now count the ways to put Joe and Jim together on the floor. There are 10C2 ways to choose the door volunteers. After that, there are 8C2 ways to choose the remaining two floor volunteers, for a total of
10C2 * 8C2 = 45 * 28 = 1260.
Now count the ways to put them together as floaters. (I assume that "floaters" are considered to be together, in spite of the name.) There are 10C2 ways to choose the door volunteers and then 8C4 ways to choose the floor volunteers, for a total of
10C2 * 8C4 = 45 * 70 = 3150.
Therefore the final answer is
13860 - 210 - 1260 - 3150 = 9240.
If you want a slightly fancier argument, you can try this. Suppose Joe is given a random assignment first, and then Jim second. There is a 2/12 * 1/11 chance that they will both be assigned to the door. (For Joe, there are two spots at the door available, out of 12 total; if Joe is assigned to the door, then Jim has 11 remaining spots, 1 of which is the door.) Similarly, there is a 4/12 * 3/11 chance that they are both assigned to the floor, and a 6/12 * 5/11 chance that they're both assigned to be floaters. Therefore the total probability that they are assigned to the same job is
12C2 * 10C4 = 66 * 210 = 13860.
Now count the ways to put Joe and Jim together at the door. If you do that, then there are 10 volunteers remaining, so there are 10C4 ways to assign the floor volunteers, and then the rest are floaters. Therefore there are 10C4 = 210 arrangements with Joe and Jim together at the door.
Now count the ways to put Joe and Jim together on the floor. There are 10C2 ways to choose the door volunteers. After that, there are 8C2 ways to choose the remaining two floor volunteers, for a total of
10C2 * 8C2 = 45 * 28 = 1260.
Now count the ways to put them together as floaters. (I assume that "floaters" are considered to be together, in spite of the name.) There are 10C2 ways to choose the door volunteers and then 8C4 ways to choose the floor volunteers, for a total of
10C2 * 8C4 = 45 * 70 = 3150.
Therefore the final answer is
13860 - 210 - 1260 - 3150 = 9240.
If you want a slightly fancier argument, you can try this. Suppose Joe is given a random assignment first, and then Jim second. There is a 2/12 * 1/11 chance that they will both be assigned to the door. (For Joe, there are two spots at the door available, out of 12 total; if Joe is assigned to the door, then Jim has 11 remaining spots, 1 of which is the door.) Similarly, there is a 4/12 * 3/11 chance that they are both assigned to the floor, and a 6/12 * 5/11 chance that they're both assigned to be floaters. Therefore the total probability that they are assigned to the same job is
12
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