Donuts Part2 (but this can be done with logic)
Alice has 3 boxes of donuts. After the party, only 2 remained in one of the boxes, the glazed one and the chocolate, and the other ones are empty. She's thinking about sharing them with Clara and Matt. She could simply hand them out later on when she meets with them, but since she's taking combinatorics, she wondered in how many ways could she distribute the donuts among the boxes, where one box would go to Clara, another to Matt and the last 1 for herself. She won't do the following in real life, but she still wonders in how many ways she could do that assuming that boxes could be left empty. For example, she could keep of all the donuts and thus leaving them in just one box. All of the donuts will have to go into at least 1 box (not one left on the table or something).
So, in how many ways can she do this?
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1 solution
Let G, C and S represent each donut. Then we have
GC on box1
GC on box2
GC on box3
G on box1 and C on box2
G on box1 and C on box3
G on box2 and C on box1
G on box2 and C on box3
G on box3 and C on box1
G on box3 and C on box2
There’s a shortcut: the glazed donut can go in any of the 3 boxes and no matter which one we choose, the chocolate donut can go in either of the 3. By the multiplication principle, there are 3 × 3 possibilities.
This is similar to choosing 2 subsets of { 1 , 2 } that do not overlap.