Suppose there are 20 items and 12 bags. How many ways can the items be put into the bags if the items are distinct.
Could someone offer me a little direction on this problem. Thanks
Suppose there are 20 items and 12 bags. How many ways can the items be put into the bags if the items are distinct.
Could someone offer me a little direction on this problem. Thanks
First, why do you assume that the bags are distinct? That is not part of the statement.
If the items are distinct and the bags are distinct, then that is simply counting the number of functions from a set of twenty to a set of twelve:
$\displaystyle 12^{20}$.
But that answer means some bags may be empty.
If you require no empty bags, then that is simply counting the number of surjections from a set of twenty to a set of twelve:
$\displaystyle \sum\limits_{k = 0}^{12} {\left( { - 1} \right)^k \binom{ 12}{k} (12 - k)^{20} }$.
BUT, I don't think it means that the bags are distinct.
I'm pretty sure my previous post was incorrect, so how would I solve this problem if the items are identical. In other words:
Suppose there are 20 items and 12 bags. How many ways can the items be put into the bags if the items are identical?
Would this be a problem of placing indistinguishable items into distinguishable boxes?
I used this theorem from my book regarding "Indistinguishable objects and distinguishable boxes": C(n + r - 1, r).
We have 20 indistinguishable books going in 12 distinguishable bags. Let n = 12 and r = 20. We have C(12 + 20 - 1, 20) = C(31, 20) different ways place the items.
I'm not sure if this is correct or not. Suggestions?
You may find this webpage useful.
If the 20 items are identical but the bags are distinct go ahead and put one item in each bag. Now none is empty. There are eight items left, so:
$\displaystyle \binom{8+12-1}{8}$.