Counting: 2 dozen different donuts are given to 24 kids

combinatoricsdiscrete mathematics

$2$ dozen donuts are given out to $24$ kids so that each kid gets one donut.

1. How many ways are there to distribute the donuts if the donuts are all different?

Answer: $24!$

Why couldn't we use the product rule here and say 24 donuts x 24 kids = total different outcomes. Got confused and wondering how this is different than say having $3$ shirts and $2$ pants, then you can make $6$ different outfits.

2. How many ways are here to distribute the donuts if there are 4 varieties of donuts and exactly six of each variety

Answer: $24! / 6!6!6!6!$

I've haven't seen an example like this in my textbook, is there a formal method used here? A simple explanation of the reasoning behind this appreciated.

UPDATE: For #2, I just learned the concept in class. The method used is called "Permutation with Repetitions" for anyone learning it. Common example is distinguishable balls placed into distinct bins

Best Answer

First part:

Consider instead the different ways of order 24 kids in line so as to distribute the donuts as they get unpacked.

Second part:

Consider the equivalent calculation of:

  1. Choosing, among 24 kids, 6 of them to give them the first kind of donuts, i.e. $24\choose 6$.
  2. Of the 18 kids still without donuts, we give out 6 donuts of some other kind out of the 4 kinds available to 6 other kids, i.e. $18 \choose 6$.
  3. Next we have $12$ kids to choose from for the 6 donuts of the third kind, i.e. $12\choose 6$.
  4. And we end up with one single way of giving out the last kind, i.e. $6\choose 6$.

Therefore the final calculus is:

$${24\choose 6} {18\choose 6} {12\choose 6} {6\choose 6}= \frac{24!}{ 6!6!6!6!}$$

Related Question