For non-negative integers a and b where \(a \ge 2\) and \(b \ge 2\) simplify:

\(\[\binom{a + b}{2} - \binom{a}{2} - \binom{b}{2}.\]\)

Guest Jun 10, 2023

#1**0 **

By the Pascal Identity, we have

C(a + b, 2) = C(a, 2) + C(a - 1, 1) C(b, 1) + C(a - 2, 2)

Substituting this into the expression, we get

C(a + b, 2) - C(a, 2) - C(b, 2) = C(a - 1, 1) C(b, 1) + C(a - 2, 2)

Since a≥2 and b≥2, we know that a−1≥1 and a−2≥0. Therefore, C(a−1,1)≥1 and C(a−2,2)≥0. This means that the expression is always nonnegative.

We can also simplify the expression by expanding the terms. We get

C(a + b, 2) - C(a, 2) - C(b, 2) = \frac{(a + b)(a + b - 1)}{2} - \frac{a(a - 1)}{2} - \frac{b(b - 1)}{2}

This simplifies to

(ab + a + b)/2.

This expression is always nonnegative, and it is equal to zero if and only if a=b.

Guest Jun 10, 2023