We need to find the coefficient of $x^6$ in the binomial expansion of $(x+2)^{10}$.

1. Problem Description

We need to find the coefficient of $x^6$ in the binomial expansion of $(x+2)^{10}$.

2. Solution Steps

The binomial theorem states that for any non-negative integer $n$ and any real numbers $a$ and $b$:

$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.

In our problem, we have $a = x$, $b = 2$, and $n = 10$. We are looking for the term with $x^6$, which means we want $n-k = 6$, so $10-k = 6$, which gives $k = 4$.

The term with $x^6$ is given by:

$\binom{10}{4} x^{10-4} 2^4 = \binom{10}{4} x^6 2^4$.

We need to compute $\binom{10}{4}$:

$\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$.

We also need to compute $2^4 = 16$.

Therefore, the term with $x^6$ is $210 \times x^6 \times 16 = 3360x^6$.

The coefficient of $x^6$ is $210 \times 16 = 3360$.

3. Final Answer

The coefficient of $x^6$ in the binomial expansion of $(x+2)^{10}$ is 3360.