[THOUGHT] We need to compute $\mathbb{E}[N^4]$ for $N \sim \text{Binomial}(n=20, p=0.5)$, then $K = 1000 \cdot \mathbb{E}[N^4]$.

We can compute $\mathbb{E}[N^4]$ using the known formula for moments of binomial distribution, or directly via summation:

We can compute this exactly using Python with fractions to get an exact rational number, then multiply by 1000 and get the integer K.

Alternatively, we can use sympy to compute the expectation symbolically.

Let's compute using Python with sympy's binomial and rational arithmetic.

Plan:

1. Use sympy to sum over k from 0 to 20: k^4 * C(20,k) * (1/2)^20.
2. Multiply by 1000.
3. Convert to integer (since result should be integer).

We'll do that now.

[/PYTHON] [PYTHON] import sympy as sp

n = 20 p = sp.Rational(1, 2)

Compute E[N^4] exactly

E_N4 = sum(k4 * sp.binomial(n, k) * pk * (1-p)**(n-k) for k in range(n+1)) print("E[N^4] =", E_N4) print("E[N^4] as rational:", E_N4.simplify())

K = 1000 * E_N4 print("K =", K) print("K as integer:", int(K)) [/PYTHON]

[THOUGHT] From the Python output, we have $\mathbb{E}[N^4] = \frac{26145}{2}$, so $K = 1000 \cdot \frac{26145}{2} = 500 \cdot 26145 = 13072500$. The integer value is 13072500.

$\boxed{13072500}$