Online Course Discussion Forum
Math Challenge II-A Combinatorics
Lecture 1, Problem 1.15
The answer says that it is C. Wouldn't the answer be A if the slices aren't the same size? If they aren't the same size, they are different so the first slice could go to 6 different people, the second slice could go to 6 different people, etc. The answer is $6\cdot 6\cdot 6\cdots 6$, or $6^{10}$.
If the slices are the same size, then it is basically throwing 10 balls into 6 different boxes. Using the stars and bars trick, the answer is $\dbinom{6+10-1}{6-1}$, or $\dbinom{15}{5}$, which is $\dfrac{15\cdot 14 \cdot 13 \cdot 12 \cdot 11}{5\cdot 4 \cdot 3 \cdot 2 \cdot 1}$. The answer is none.
The slices of cake are different (think about an actual cake, even if you try to cut exactly the same, some slices will have different frosting, be slightly bigger/smaller, etc.).
The problem with an answer of $6^{10}$ is that each person is choosing "a", meaning one, slice of cake. Therefore the first person has $10$ choices, the second $9$, etc, giving answer choice (C). If you wanted to do people choosing slices of cake, you'd first need to choose which $6$ of the $10$ slices are used (so order doesn't matter), and then give out one slice to each person (be careful here, as then this is with no repetitions). This would give $\binom{10}{6} \cdot 6!$ which also simplifies to (C).
Social networks