-80
10
40
90
Correct answer is C
x\(^3\)y\(^2\) in (x-2y)\(^5\)
n = 5, r = 3, p = x, q = -2y
5C\(_3\) * x\(^3\) -2y\(^2\)
5C\(_3\) = \(\frac{5!}{[5-3]!3!}\)
\(\frac{5*4*3!}{2! 3!}\) → \(\frac{5*4}{2}\)
5C\(_3\) = 10
: 5C\(_3\) * x\(^3\) -2y\(^2\) = 10 * x\(^3\) 4y\(^2\)
40x\(^3\)y\(^2\)
the coefficient is 40