The value of $$\sum_{r=0}^{20} {}^{50-r}C_6$$ is equal to:

We have to find the numerical value of the finite sum

$$\sum_{r=0}^{20} {^{\,50-r}\!C_{\,6}}.$$

The upper index of the combination depends on $$r$$, so first we replace that dependent index by a single symbol that increases in the usual (left-to-right) direction. Put

$$k = 50 - r.$$

When $$r = 0$$, we obtain $$k = 50$$, and when $$r = 20$$, we obtain $$k = 30$$. As $$r$$ runs from $$0$$ to $$20$$, $$k$$ therefore runs from $$50$$ down to $$30$$. Because addition is commutative, the order does not matter, so we may rewrite the sum as

$$\sum_{r=0}^{20} {^{\,50-r}\!C_{\,6}} \;=\; \sum_{k=30}^{50} {^{\,k}\!C_{\,6}}.$$

Now we recall a standard combinatorial identity, usually proved by induction or by Pascal’s rule:

$$\sum_{k=m}^{n} {^{\,k}\!C_{\,m}} \;=\; {^{\,n+1}\!C_{\,m+1}} \quad\text{for integers } n \ge m \ge 0.$$

In our case $$m = 6$$, so if the lower limit were $$6$$ instead of $$30$$, we could apply the identity directly. We therefore write the desired sum as the difference of two larger sums whose lower limit is $$6$$:

$$\sum_{k=30}^{50} {^{\,k}\!C_{\,6}} \;=\; \left(\sum_{k=6}^{50} {^{\,k}\!C_{\,6}}\right) \;-\; \left(\sum_{k=6}^{29} {^{\,k}\!C_{\,6}}\right).$$

By the quoted identity, we evaluate each parenthesis separately.

First parenthesis (upper limit $$50$$):

$$\sum_{k=6}^{50} {^{\,k}\!C_{\,6}} = {^{\,50+1}\!C_{\,6+1}} = {^{\,51}\!C_{\,7}}.$$

Second parenthesis (upper limit $$29$$):

$$\sum_{k=6}^{29} {^{\,k}\!C_{\,6}} = {^{\,29+1}\!C_{\,6+1}} = {^{\,30}\!C_{\,7}}.$$

Substituting these two evaluations back, we get

$$\sum_{k=30}^{50} {^{\,k}\!C_{\,6}} = {^{\,51}\!C_{\,7}} \;-\; {^{\,30}\!C_{\,7}}.$$

Hence, the original sum equals $${}^{51}C_7 - {}^{30}C_7$$, which coincides with Option A.

Hence, the correct answer is Option A.