Sum of Squares of Binomial Coefficients

$$\left( \sum_{r=0}^{n} a_r \right)^2 = \sum_{r=0}^{n} a_r^2 + 2 \sum_{0 \le i < j \le n} a_i a_j$$
Let $$a_r = \binom{10}{r}$$ and $$n = 10$$. Substituting these into the identity gives:
$$\left( \sum_{r=0}^{10} \binom{10}{r} \right)^2 = \sum_{r=0}^{10} \binom{10}{r}^2 + 2 \sum_{0 \le i < j \le 10} \binom{10}{i} \binom{10}{j}$$
We know two standard binomial properties:
1. Sum of coefficients: $$\sum_{r=0}^{10} \binom{10}{r} = 2^{10}$$
2. Sum of squares of coefficients: $$\sum_{r=0}^{10} \binom{10}{r}^2 = \binom{20}{10}$$
Substitute these values back into our main equation:
$$(2^{10})^2 = \binom{20}{10} + 2S$$
$$2^{20} = \binom{20}{10} + 2S$$
Isolate the target sum ($$S$$):
$$2S = 2^{20} - \binom{20}{10}$$
$$S = \frac{2^{20} - \binom{20}{10}}{2} = 2^{19} - \frac{1}{2}\binom{20}{10}$$
S + $$\frac{1}{2}\binom{20}{10} =2^{19}$$