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Math Challenge II-A Combinatorics

 
 
LiangNeo的头像
Math Challenge II-A Combinatorics
LiangNeo - 2019年09月1日 Sunday 15:14
 

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.



 
ProfessorAreteem的头像
Re: Math Challenge II-A Combinatorics
ProfessorAreteem - 2019年09月4日 Wednesday 13:11
 

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).