The problem asks us to find the coefficient of $x^4$ in the expansion of $(1-2x)^6$.

1. Problem Description

The problem asks us to find the coefficient of $x^4$ in the expansion of $(1-2x)^6$.

2. Solution Steps

We will use the binomial theorem to expand $(1-2x)^6$. The binomial theorem states that for any non-negative integer $n$ and any real numbers $a$ and $b$:

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

In our case, $a=1$, $b=-2x$, and $n=6$. We want the term with $x^4$, so we need $k=4$. Thus, we have

${6 \choose 4} (1)^{6-4} (-2x)^4 = {6 \choose 4} (1)^2 (-2x)^4$

We first calculate the binomial coefficient:

${6 \choose 4} = \frac{6!}{4! (6-4)!} = \frac{6!}{4! 2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) (2 \times 1)} = \frac{6 \times 5}{2 \times 1} = 15$

Next, we simplify the term:

$(1)^2 = 1$

$(-2x)^4 = (-2)^4 x^4 = 16 x^4$

Therefore, the term with $x^4$ is:

$15 \times 1 \times 16x^4 = 240 x^4$

The coefficient of $x^4$ is

2

4

0.

3. Final Answer

The coefficient of $x^4$ in the expression of $(1-2x)^6$ is 240.