Combinations: Prize Distribution

[ChaoticLlama]

As winners of a competition, 3 students must decide how to distribute 12 different prizes amongst themselves. In how many ways can they do this if each person gets 4 prizes but nobody can get BOTH the CD player & the Palm Pilot (assume the CD player and Palm Pilot are 2 of the 12 prizes)

This question is causing me confusion. I have been very good at this unit however this review question for a test has me stumped. What I've tried is find the total number of distributions and subtracting three disjoint cases, where I force person 1 to hold both prizes, and the same with person 2 and person 3.

Total : (12C4)(8C4)(4C4)
where you start with 12 prizes, give 4 to the first guy, then you have 8 remaining to give 4 to person number 2, then 4 prizes for the 4 last spots.

Then..

I subtract 3 disjoint cases from the total to get the answer
Person 1 has both: [(2C2)(10C2)](8C4)(4C4)
Person 2 has both: (12C4)[(2C2)(6C2)](4C4)
Person 3 has both: (12C4)(8C4)[(2C2)(2C2)]

However, there is an error in my logic, because the case I created which forces person 3 to have both prizes, is equal to the total number of distributions.

Thanks for you help.

 

[Gene]

I would hand out the CD and Palm first. (3*2) ways.
I would then hand out 1 of the 10 to the 3rd person (10*1) ways.
Finally the remaining 9, 3 at a time. (9C3)*(6C3)*(3C3) ways.
I don't recall that last formula but see no flaw in your reasoning about it. If it is correct the answer should be
6*10*(9C3)*(6C3)*(3C3)

 

[pka]

It maybe too late to worry with an answer, but this old Prof sees a chance.
If we gave out the 12 prizes without restrictions there would be \(\displaystyle \left( {_{12} C_4 } \right)\left( {_8 C_4 } \right)\left( {_4 C_4 } \right) = 34,650\) ways to do it.
Look at the following: \(\displaystyle \left( 6 \right)\left( {10} \right)\left( {_9 C_3 } \right)\left( {_6 C_3 } \right)\left( {_3 C_3 } \right) = 100,800\)!
In other words, the proposed answer exceeds to total possible by some bit.
Over counting has taken place.

I suggest that the correct answer is: \(\displaystyle \left( 6 \right)\left( {_{10} C_4 } \right)\left( {_6 C_3 } \right)\left( {_3 C_3 } \right) = 25,200\).

To see this, consider the possibility that the students are Adams, Brown, and Clay.
A string of four A’s, four B’s, and four C’s can be arranged in \(\displaystyle \frac{{12!}}{{\left( {4!} \right)^3 }}\) ways. So that is the total possible to assign the 12 prizes.
There are 6 ways to assign the two ‘big’ prizes.
Say Brown is left out then we assign four B’s, three A’s and three C’s to the remaining ten prizes: \(\displaystyle 6\frac{{10!}}{{\left( {3!} \right)^2 \left( {4!} \right)}} = 25,200\).

 

[ChaoticLlama]

Thanks pka, that is the correct answer

Another method that I found if you are curious..

(12C4)(8C4)(4C4) - ((2C2)*3)(10C2)(8C4)(4C4) = 25,200

where (2C2) are the two prizes, and you multiply by three to assign it to differnt people.

 

[Gene]

Sorry gang. I was so sure of my logic that I didn't bother to run the numbers. I'm still not sure where I overcounted. It must be in the 10 to the third and the 3/9 to each but so far I can't see it. Good catch
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Gene