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Mock AIME 5 Pre 2005 Problems/Problem 4

Revision as of 00:05, 15 February 2024 by Abbywong (talk | contribs) (Created page with "Let the number of balls in the 8 boxes be <math>a_1,a_2,a_3,...,a_8</math>. Thus, we have <math>a_1+a_2+a_3+...+a_8=16</math>, where <math>a_1,a_2,a_3,...,a_8\geq1</math>. By...")
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Let the number of balls in the 8 boxes be $a_1,a_2,a_3,...,a_8$. Thus, we have $a_1+a_2+a_3+...+a_8=16$, where $a_1,a_2,a_3,...,a_8\geq1$. By stars and bars, the number of ways this is possible is $\binom{16-1}{8-1}=\binom{15}{7}=6435$, and since $15+7=22$, the answer is $6435\mod22=\boxed{11}$. ~AbbyWong

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