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Difference between revisions of "2025 SSMO Speed Round Problems/Problem 3"

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<cmath>\tbinom{2}{2} + \tbinom{3}{2} + \cdots + \tbinom{14}{2}.</cmath>
 
<cmath>\tbinom{2}{2} + \tbinom{3}{2} + \cdots + \tbinom{14}{2}.</cmath>
 
By the hockey stick identity, this is equal to <math>\tbinom{15}{3} = \boxed{455}</math>.
 
By the hockey stick identity, this is equal to <math>\tbinom{15}{3} = \boxed{455}</math>.
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~Sedro

Latest revision as of 15:13, 9 September 2025

Problem

Anna is buying different types of cheese from the local supermarket. Let $x,$ $y,$ and $z$ be the number of pieces of blue, cheddar, and mozzarella cheese, respectively, that Anna buys. She can buy any nonnegative integer number of each type, but the total number of pieces must be at most 12. How many different combinations $(x, y, z)$ of cheese can Anna buy? (Anna is allowed to buy 0 pieces of cheese.)

Solution

We seek the number of ordered pairs $(x,y,z)$ of nonnegative integers such that $0\le x+y+z\le 12$. Let $0\le n \le 12$ be some integer. By stars and bars, there are $\tbinom{n+2}{2}$ ordered pairs $(x,y,z)$ of nonnegative integers satisfying $x+y+z = n$. Thus, the answer is the sum \[\tbinom{2}{2} + \tbinom{3}{2} + \cdots + \tbinom{14}{2}.\] By the hockey stick identity, this is equal to $\tbinom{15}{3} = \boxed{455}$.

~Sedro

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