It looks like you've got some pieces of an algebraic expression and its expansion, but the notation is somewhat unclear. Let's try to work through the expansion of [tex]\((2x - 3y)^{25}\)[/tex] step by step to find the coefficient of [tex]\(x^{12}y^{13}\)[/tex].
We're expanding [tex]\((2x - 3y)^{25}\)[/tex] using the binomial theorem, which states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In our case, [tex]\(a = 2x\)[/tex], [tex]\(b = -3y\)[/tex], and [tex]\(n = 25\)[/tex]. Thus, the expansion is:
[tex]\[
(2x - 3y)^{25} = \sum_{k=0}^{25} \binom{25}{k} (2x)^{25-k} (-3y)^k
\][/tex]
We want the coefficient of the term [tex]\(x^{12} y^{13}\)[/tex]. This means we need to find [tex]\(k\)[/tex] such that the power of [tex]\(x\)[/tex] is 12 and the power of [tex]\(y\)[/tex] is 13.
Since [tex]\((2x)^{25-k} \)[/tex] and [tex]\( (-3y)^k \)[/tex], make:
[tex]\[
(2x)^{25-k} = 2^{25-k} x^{25-k}
\][/tex]
[tex]\[
(-3y)^k = (-3)^k y^k
\][/tex]
We want:
[tex]\[
25 - k = 12 \quad \text{(for the power of } x \text{)}
\][/tex]
[tex]\[
k = 13 \quad \text{(for the power of } y \text{)}
\][/tex]
The value of [tex]\(k\)[/tex] is 13. Now, substitute [tex]\(k\)[/tex] back into the binomial coefficient and the terms:
[tex]\[
\binom{25}{13} (2)^{25-13} (-3)^{13} x^{12} y^{13}
\][/tex]
Calculate the coefficient:
[tex]\[
\binom{25}{13} = \frac{25!}{13! \cdot 12!}
\][/tex]
Therefore:
[tex]\[
2^{12}
\][/tex]
[tex]\[
(-3)^{13}
\][/tex]
Putting it together, the coefficient is:
[tex]\[
\binom{25}{13} \cdot 2^{12} \cdot (-3)^{13}
\][/tex]
[tex]\[
= \frac{25!}{13! \cdot 12!} \cdot 2^{12} \cdot (-3)^{13}
\][/tex]
The final expression for the coefficient of [tex]\(x^{12} y^{13}\)[/tex] is:
[tex]\[
\frac{25!}{13! \cdot 12!} \cdot 2^{12} \cdot (-3)^{13}
\][/tex]
