Assume that no one can win more than one prize.

If the prizes were all different, then we have the case that order matters and repeats are allowed, meaning there are $P(51, 10)$ ways of handing out the prizes.

But the prizes are not all different.

I know that I need to:

  1. Define an equivalence relation. Should I define the relation to be two prizes are the same given they are both pencils or both cars, or should I define two prizes to be the same provided that they are given to the same person?

  2. Count the equivalence classes. I will need to know the answer to #1 to do this.

EDIT: My attempt at the solution.

Define two prizes to be the same provided that they are both pencils or both cars. This is an equivalence relation on our $P(51, 10)$ ways to hand out the prizes. There are $7!$ ways to hand out the pencils, and $3!$ ways to hand out the cars, so there are $\frac{P(51, 10)}{7!3!}$ equivalence classes. Thus there are $\frac{P(51, 10)}{7!3!}$ ways to hand out the prizes.

Is this solution correct?

  • 2
    $\begingroup$ The solution is correct. I would do it differently. Assume the cars are identical and the pencils are identical and that a pencil is different from a car. There are $\binom{51}{3}$ ways to decide who gets the cars, For each of these ways, there are $\dots$. $\endgroup$ – André Nicolas Mar 2 '16 at 1:46

Your answer is correct. Since each person gets at most one prize, we simply need to choose which people receive the cars and which of the remaining people receive the pencils. Three of the fifty-one people receive a car and seven of the remaining forty-eight people receive a pencil. Hence, the number of ways the prizes can be distributed is $$\binom{51}{3}\binom{48}{7}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.