Grandma has 8 grandchildren, and 4 different types of popsicles:
- 6 Vanilla popsicles
- 6 Strawberry popsicles
- 5 Banana popsicles
- 3 Chokolate popsicles
This morning, all of her grandchildren came together and asked for one popsicle each (every grandchild asked for a particular flavor). What is the total number of different sets of requests that Grandma can fulfill?
I think this is related to the Inclusion-Exclusion principle because it was taught in the same class. Can you help me solve it?
I did reach the following sum, but I imagine the question's author had something simpler in mind...
$ E(0) = W(0)-W(1) = 3^8 - 4\cdot C(8,8)-4\cdot 3\cdot C(7, 8)-2\cdot C(6, 8)\cdot 3^2-C(5, 8)\cdot 3^3 - C(4, 8)\cdot 3^4$
I don't quite understand how you arrived at your result, so I'll try to sketch my solution. As you tagged the question as homework, there are no computations - just the idea.
There are at first $4$ possibilities to consider, depending how many chocolate popsicles the kids want: $0,1,2$ or $3$. Say $A(i)$ is the number of requests that have $i$ chocolate popsicles chosen. Let $T$ be the desired total. Then clearly:
$$T = A(0) + A(1) + A(2) + A(3)$$
as the possibilities are mutually exclusive and exhaustive.
$A(3)$ is easy: $3$ kids picked chocolate, and the remaining $5$ are choosing amongst the other three flavours. Note that they can all choose the same flavour.
$A(2)$ is just as easy: you have $6$ kids left to consider. Assume first that there are $6$ banana popsicles as well. Count all possible choices, then discount the one case when all six kids pick the banana popsicle.
$A(1)$ and $A(0)$: this would be hard to compute as directly as the previous two. Try it this way: counting $A(1)$ is equivalent to the problem:
Which is just the original problem minus one child and the chocolate popsicles. You approach it as above. Let $B(i)$ be the number of requests where $i$ banana popsicles are picked.
Then you have:
$$A(1) = B(0) + B(1) + B(2) + B(3) + B(4) + B(5)$$
All these $B(i)$ are now easy to compute. A similar calculation works for $A(0)$.