To find the coefficient of [tex]\(x^{72}y^2\)[/tex] in the expansion of [tex]\((x-y)^{74}\)[/tex], we will use the Binomial Theorem.
The Binomial Theorem states that for any positive integer [tex]\(n\)[/tex], the expression [tex]\((x+y)^n\)[/tex] can be expanded as:
[tex]\[
(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\][/tex]
Where [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient, defined as:
[tex]\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\][/tex]
In our case, the expression is [tex]\((x-y)^{74}\)[/tex]. To account for the negative sign, we modify our Binomial Theorem application to:
[tex]\[
(x-y)^{74} = \sum_{k=0}^{74} \binom{74}{k} x^{74-k} (-y)^k
\][/tex]
We are looking for the term involving [tex]\(x^{72} y^2\)[/tex]. This corresponds to [tex]\(n=74\)[/tex], [tex]\(k=2\)[/tex], since the exponent [tex]\(n-k\)[/tex] for [tex]\(x\)[/tex] is [tex]\(74-2=72\)[/tex] and the exponent for [tex]\(y\)[/tex] is [tex]\(2\)[/tex].
The general term in the expansion of [tex]\((x-y)^{74}\)[/tex] can be written as:
[tex]\[
\binom{74}{k} x^{74-k} (-y)^k
\][/tex]
For [tex]\(k=2\)[/tex], the term becomes:
[tex]\[
\binom{74}{2} x^{74-2} (-y)^2 = \binom{74}{2} x^{72} (y^2)
\][/tex]
We need to calculate [tex]\(\binom{74}{2}\)[/tex] and determine the effect of [tex]\((-y)^2\)[/tex]:
1. Calculate the binomial coefficient [tex]\(\binom{74}{2}\)[/tex]:
[tex]\[
\binom{74}{2} = \frac{74!}{2!(74-2)!} = \frac{74 \times 73}{2 \times 1} = 2701
\][/tex]
2. Determine the contribution of [tex]\((-y)^2\)[/tex]:
[tex]\[
(-y)^2 = y^2 \quad \text{(since the square of a negative number is positive)}
\][/tex]
Thus, the coefficient of [tex]\(x^{72} y^2\)[/tex] in [tex]\((x-y)^{74}\)[/tex] is:
[tex]\[
\binom{74}{2} = 2701
\][/tex]
Therefore, the exact coefficient of [tex]\(x^{72} y^2\)[/tex] in the expansion of [tex]\((x-y)^{74}\)[/tex] is [tex]\( \boxed{2701} \)[/tex].
