To find the coefficient of the \(x^7 y^3\) term in the expansion of \((x-2y)^{10}\), we can use the binomial theorem. The binomial theorem states that:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this problem, we have \(a = x\), \(b = -2y\), and \(n = 10\). We are interested in the term where the power of \(y\) is 3, which corresponds to \(k = 3\).
The general term in the expansion is given by:
[tex]\[
\binom{n}{k} x^{n-k} (-2y)^k
\][/tex]
Substituting \(n = 10\), \(k = 3\), and \(n-k = 7\), we have:
[tex]\[
\binom{10}{3} x^{7} (-2y)^3
\][/tex]
Now we need to calculate each component of this term:
1. Calculate the binomial coefficient \(\binom{10}{3}\):
[tex]\[
\binom{10}{3} = \frac{10!}{3! (10-3)!} = \frac{10!}{3! 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\][/tex]
2. Determine the powers of \(x\) and \(y\):
Since we are dealing with \(k = 3\),
- The power of \(x\) is \(10 - 3 = 7\).
- The power of \(y\) is \(3\).
3. Calculate the coefficient for \((-2y)^3\):
[tex]\[
(-2y)^3 = (-2)^3 y^3 = -8 y^3
\][/tex]
4. Multiply the binomial coefficient by the coefficient of \((-2y)^3\):
[tex]\[
120 \times (-8) = -960
\][/tex]
Thus, the coefficient of the \(x^7 y^3\) term in the expansion of \((x-2y)^{10}\) is \(-960\).
So, the correct answer is:
[tex]\[
\boxed{-960}
\][/tex]
