Correct Answer: Option C

Explanation

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

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