Jon S. answered 11/04/20
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C(5,0)(5x)^5(6y)^0 - C(5,1)(5x)^4(6y)^1 + C(5,2)(5x)^3(6y)^2 - C(5,3)(5x)^2(6y)^3 + C(5,4)(5x)^1(6y)^4 - C(5,5)(5x)^0(6y)^5
where C(5,x) is number of combinations of 5 taken 2 at a time