Source: 2018 Canadian Open Math Challenge Part B Problem 4 Determine the number of $5$-tuples of integers $(x_1,x_2,x_3,x_4,x_5)$ such that $\text{(a)}$ $x_i\ge i$ for $1\le i \le 5$; $\text{(b)}$ $\sum_{i=1}^5 x_i = 25$.
Problem
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Tags: Comc, 2018 COMC
NikoIsLife
06.12.2018 17:15
Let $y_i=x_i+i$. Then, we are just finding the number of nonnegative integer solutions to:
$$y_1+y_2+y_3+y_4+y_5=10$$By Stars and Bars, this is just $\binom{14}4=\boxed{1001}$.
ahwawm
06.12.2018 18:02
NikoIsLife wrote:
Let $y_i=x_i+i$. Then, we are just finding the number of nonnegative integer solutions to:
$$y_1+y_2+y_3+y_4+y_5=10$$By Stars and Bars, this is just $\binom{14}4=\boxed{1001}$.
Stars and Bars?? Could you explain??
enthusiast101
06.12.2018 18:25
Total solutions to $x_1+x_2+x_4+x_4+x_5=25$ is $\binom{25+(5-1)}{4}=\binom{29}{4}=23751$. Following NikoIsLife's solution, no. Of solutions to $\sum_{k=1}^{5}x_k=10$ is $\binom{10-(5-1)}{5-1}=\binom{14}{4}=1001$.