Can someone please help me understand the following three variations of a problem…
Here are the three problems along with some of my reasoning/questions:
1) How many ways to put 20 different (distinct) chocolates into a red bag and a green bag so that each bag contains 10 chocolates?
I'm thinking I can think of this as a set (call this set $A$) of 20 distinct integers and the different ways I can put these distinct elements into two distinct sets (call these sets $B$ and $C$).
My answer to this is: ${20 \choose 10} \cdot {10 \choose 10}$. That is select 10 elements for my first set $B$ and 10 elements out of the remaining elements for my second set $C$. This one seems pretty straightforward.
2) How many ways to put 20 different (distinct) chocolates into two identical blue bags so that each bag contains 10 chocolates?
Conceptually this is a bit trickier. I'm thinking this is analogous to a set (call this set $A$) of 20 distinct integers and the ways I can put these distinct elements into two identical sets (both called $B$).
My answer to this is: $\frac{20\choose 10}{2!}$. That is, I'm thinking of this as the number of ways to create two sets of subsets, but since one set is identical to the other I'm dividing out the over-counted subsets. My question here though is how is this possibly less than the ways I can create 10 element subsets from a 20 element set? I'm dealing with two sets… although identical it doesn't make intuitive sense that this somehow reduces the number of combinations that would be obtained from a single set.
3) How many ways to put 20 identical chocolates into two identical blue bags so that each bag contains 10 chocolates?
This one I'm not sure on.
Best Answer
Your answer to #1 is correct, and so is your answer to #2.
Here is one way to justify the answer to #2:
If we put the chocolates into two identical blue bags, and then color one blue bag red and the other one green, then we will get the result in #1.
Since there are $2!$ ways to color the bags, the answer to #1 is $2!$ times the answer to #2.
The answer to #3 is 1, since the chocolates and the bags are identical.