If $$\sum_{k=1}^{10} K^2 (10_{C_{K}})^{2} = 22000L$$, then L is equal to _____

We need to find $$L$$ such that $$\displaystyle\sum_{k=1}^{10} k^2 \binom{10}{k}^2 = 22000L$$.

We begin by simplifying $$k^2\binom{10}{k}$$ using the standard identity $$k\binom{n}{k} = n\binom{n-1}{k-1}$$. Applying this once gives $$k\binom{10}{k} = 10\binom{9}{k-1}$$, and then multiplying by $$k$$ and writing $$k = (k-1) + 1$$:

$$k^2\binom{10}{k} = k \cdot 10\binom{9}{k-1} = 10\bigl[(k-1) + 1\bigr]\binom{9}{k-1} = 10(k-1)\binom{9}{k-1} + 10\binom{9}{k-1}.$$

Now applying the identity again to the first term: $$(k-1)\binom{9}{k-1} = 9\binom{8}{k-2}$$. So we obtain $$k^2\binom{10}{k} = 90\binom{8}{k-2} + 10\binom{9}{k-1}$$.

Substituting into the sum and using a second copy of $$\binom{10}{k}$$:

$$\sum_{k=1}^{10} k^2\binom{10}{k}^2 = 90\sum_{k=2}^{10}\binom{8}{k-2}\binom{10}{k} + 10\sum_{k=1}^{10}\binom{9}{k-1}\binom{10}{k}.$$

For the first sum, we substitute $$j = k - 2$$: $$\displaystyle\sum_{j=0}^{8}\binom{8}{j}\binom{10}{j+2}$$. Using the identity $$\binom{10}{j+2} = \binom{10}{8-j}$$ and the Vandermonde convolution $$\displaystyle\sum_{j=0}^{8}\binom{8}{j}\binom{10}{8-j} = \binom{18}{8}$$. We compute $$\binom{18}{8} = 43758$$.

For the second sum, we substitute $$j = k - 1$$: $$\displaystyle\sum_{j=0}^{9}\binom{9}{j}\binom{10}{j+1}$$. Since $$\binom{10}{j+1} = \binom{10}{9-j}$$, the Vandermonde convolution gives $$\displaystyle\sum_{j=0}^{9}\binom{9}{j}\binom{10}{9-j} = \binom{19}{9} = 92378$$.

Combining: $$\displaystyle\sum_{k=1}^{10} k^2\binom{10}{k}^2 = 90 \times 43758 + 10 \times 92378 = 3{,}938{,}220 + 923{,}780 = 4{,}862{,}000$$.

Setting $$4{,}862{,}000 = 22000L$$, we get $$L = \frac{4{,}862{,}000}{22{,}000} = 221$$.

Hence, the correct answer is $$\boxed{221}$$.