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Find the number of solutions to
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 \le 10
in nonnegative integers, where x_2 is even and x_5 is odd.

 May 9, 2024
 #1
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The pair \((x_2, x_5)\) can be (0, 1), (0, 3), ..., (0, 9), (2, 1), (2, 3), ..., (2, 7), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (8, 1).

 

From here, 5 pairs add up to 9, 4 pairs add up to 7, 3 pairs add up to 5, 2 pairs add up to 3, and 1 pair add up to 1.

 

If \(x_2 + x_5 = k\), then \(x_1 + x_3 + x_4 + x_6 + x_7 = 10 - k\). The number of nonnegative integer solutions of \(x_1 + x_3 + x_4 + x_6 + x_7 = N\) is \(\displaystyle \binom{N+5-1}{5-1} = \binom{N+4}4\). Then, we simply calculate the required number to be \(\displaystyle 5\binom{1 + 4}{4} + 4\binom{3 + 4}{4} + 3\binom{5 + 4}{4} + 2\binom{7+ 4}{4} + 1\binom{9+4}{4} = 1918\).

 May 10, 2024
edited by MaxWong  May 10, 2024

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