We need to find the coefficient of $x^2y^9$ in the expansion of $(3x+4y)^{11}$.

1. Problem Description

We need to find the coefficient of $x^2y^9$ in the expansion of $(3x+4y)^{11}$.

2. Solution Steps

We use the binomial theorem to find the term with $x^2y^9$ in the expansion of $(3x+4y)^{11}$.

The general term in the binomial expansion of $(a+b)^n$ is given by:

$T_{k+1} = \binom{n}{k}a^{n-k}b^k$

In our case, $a = 3x$, $b = 4y$, and $n = 11$. We want the term with $x^2y^9$.

So we need $(3x)^{11-k}(4y)^k$ to give us $x^2y^9$. This means $11-k=2$ and $k=9$. Since these are consistent, we use $k=9$.

Thus, the term we are looking for is:

$T_{9+1} = T_{10} = \binom{11}{9}(3x)^{11-9}(4y)^9 = \binom{11}{9}(3x)^2(4y)^9$

Now, we compute the binomial coefficient:

$\binom{11}{9} = \frac{11!}{9!(11-9)!} = \frac{11!}{9!2!} = \frac{11 \times 10}{2 \times 1} = 55$

Now, we compute $(3x)^2$ and $(4y)^9$:

$(3x)^2 = 9x^2$

$(4y)^9 = 4^9y^9 = 262144y^9$

So the term is:

$T_{10} = 55 \times 9x^2 \times 262144y^9 = 55 \times 9 \times 262144 x^2 y^9 = 129746880 x^2 y^9$

The coefficient of $x^2y^9$ is $55 \times 9 \times 4^9 = 55 \times 9 \times 262144 = 129746880$

3. Final Answer

The coefficient of $x^2y^9$ in the expansion of $(3x+4y)^{11}$ is $129746880$.