$$\sum_{k=0}^{6} {}^{51-k}C_3$$ is equal to

We need to evaluate $$\sum_{k=0}^{6} \binom{51-k}{3}$$.

Expanding the sum:

$$ \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3} + \binom{46}{3} + \binom{45}{3} $$

We use the Hockey Stick Identity (or Christmas Stocking Identity):

$$ \sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1} $$

Rearranging our sum in ascending order:

$$ \binom{45}{3} + \binom{46}{3} + \binom{47}{3} + \binom{48}{3} + \binom{49}{3} + \binom{50}{3} + \binom{51}{3} $$

By the Hockey Stick Identity with $$r = 3$$:

$$ \sum_{i=3}^{51} \binom{i}{3} = \binom{52}{4} $$

Our sum starts from $$i = 45$$, not $$i = 3$$. So:

$$ \sum_{i=45}^{51} \binom{i}{3} = \sum_{i=3}^{51} \binom{i}{3} - \sum_{i=3}^{44} \binom{i}{3} = \binom{52}{4} - \binom{45}{4} $$

Therefore: $$\sum_{k=0}^{6} \binom{51-k}{3} = \binom{52}{4} - \binom{45}{4}$$

This matches Option C: $${}^{52}C_4 - {}^{45}C_4$$.