|A_i| = \binom{4 + 4 - 1}{4 - 1} = \binom{7}{3} = 35

February 22, 2026

$|A_i| = \binom{4 + 4 - 1}{4 - 1} = \binom{7}{3} = 35$

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\binom{10 + 4 - 1}{4 - 1} = \binom{13}{3} = 286
Now subtract the number of solutions where at least one $ y_i > 5 $. Let’s use inclusion-exclusion.
Let $ A_i $ be the set of solutions where $ y_i \geq 6 $. Set $ z_i = y_i - 6 $, then $ z_i + y_j + \cdots = 4 $, so:
There are 4 such sets, so total for single violations: $ 4 \cdot 35 = 140 $
Now consider intersections $ A_i \cap A_j $, where two variables $ \geq 6 $. Then $ z_i + z_j + y_k + y_l = -2 $, which is impossible. So higher intersections are 0.
So total valid solutions:

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