5
$\begingroup$

Let's say I have 10 apples, 8 oranges and 7 bananas.

I want to know in how many ways I can spread them over four distinct boxes.

So my approach would be to immediately consider this as an anagram permutation problem:

AAAAAAAAAA|OOOOOOOO|BBBB|BBB

The above example would give us 10 apples on the first box, 8 oranges on the second box, 4 bananas on the third box, and then 3 bananas on the fourth box.

So the solution would be simply $$\frac{( 10 + 8 + 7 + 3)!}{10! * 8! * 7! * 3!} = \frac{28!}{10! * 8! * 7! * 3!}$$

However my teacher told me this leads to a wrong conclusion, although he didn't explain me why.

Then he went on to show me the right way to do this.

So I'd like to know not what would be the right way to solve this ( as I already know it ) but why my proposed solution doesn't work.

$\endgroup$
3
$\begingroup$

You have to chose $10$ items from the $28$ to be apples, $8$ of the rest to be oranges and $7$ of the remaining ten to be bananas. Then you have fruit in boxes. That's the formula you have given. Your problem then is that you can rearrange the fruit in the first box without changing your answer, but you don't know the number of fruit in the first box, let alone the different types, so you know there are fewer arrangements than you first thought, but have reached a bit of a dead end in the analysis.

Rather allocate the apples to boxes, then the oranges to boxes and finally the bananas to boxes. Then you keep better control over the indistinguishable things before you mix them up.

$\endgroup$
  • $\begingroup$ So this means I've actually calculated the number of ways in which I could take fruits out of the boxes, one after another. $\endgroup$ – Ruan Apr 18 '18 at 21:50
0
$\begingroup$

Your approach orders everything (all fruit, and the three partitions between the walls) together, then divides by the permutations between each type of fruit and the partitions to make them indistinct, which is understandable. The problem is, within each box, the fruit are unordered. I could have my first box contain $OAO$, and it could also contain $AOO$, and though the box contain one apple and two oranges in each case, your approach would count them as distinct.

$\endgroup$
0
$\begingroup$

I don't entirely understand where your approach is coming from, but anyways: the number $$\frac{( 10 + 8 + 7 + 3)!}{10! * 8! * 7! * 3!} = \frac{28!}{10! * 8! * 7! * 3!}$$ is the solution to the counting problem where we have 28 objects which are mutually distinct, and we have $4$ boxes into which we can fit $10,8,7,3$ objects respectively. Then the above number tells us in how many ways we can do this, if we consider configurations the same if the ordering within each box does not matter.

Clearly this problem is a little different: as Mark already notes, not all objects are indistinguisable here. Indeed, a more efficient way to solve the problem is to first distribute the $10$ apples over all boxes, then all the oranges, and after you have done so the distribution of the bananas is of course already fixed.

Hope this helps!

$\endgroup$

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.