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Combinatorial Mathematics-Indefinite Equation Problem?
Write a = x 1+x2, b = x3+x4, then a, b.

So there are 10 combinations of A and B * * *.

(3, 12), (4, 1 1), (5, 10), (6,9), (7,8), (8,7), (9,6), ( 10,5), ( 1 1,4), ( 12,3)

For x 1, x2

When a=3, (x 1, x2)=( 1, 2) or (2, 1), there are two combinations.

When a=4, (x 1, x2)=( 1, 3) or (2,2) or (3, 1), there are three combinations.

When a=5, (x 1, x2)=( 1, 4) or (2,3) or (3,2) or (4, 1), there are four combinations.

When a=6, (x 1, x2) has five combinations.

When a=7, (x 1, x2) has six combinations.

When a=8, pay attention to x 1, x2.

When a=9, note x 1, x2.

When a= 10, (x 1, x2) has three combinations.

When a= 1 1, (x 1, x2) has two combinations.

When a= 12, (x 1, x2) = (6,6), there are only 1 combinations.

When b=n, the corresponding number of combinations should be the same as when a = n.

So the total number of combinations should be

=2× 1

+3×2

+4×3

+5×4

+6×5

+5×6

+4×5

+3×4

+2×3

+ 1×2

= 140 (type)