Step 1: Understanding the Topic
This question requires the evaluation of a sum of binomial coefficients. The terms in the sum follow a specific pattern that can be simplified using a known combinatorial identity, often referred to as the Hockey-stick or Christmas stocking identity.
Step 2: Key Formula or Approach
The sum can be written in summation notation. Notice that in each term ${}^{n}C_{k}$, the upper index is always 5 greater than the lower index, i.e., $n = k+5$. The sum is:
\[
\sum_{k=0}^{6} {}^{k+5}C_{k}
\]
We will use the identity:
\[
\sum_{i=r}^{n} {i \choose r} = {n+1 \choose r+1}
\]
To apply this, we first use the symmetry property ${n \choose k} = {n \choose n-k}$.
\[
{}^{k+5}C_{k} = {}^{k+5}C_{(k+5)-k} = {}^{k+5}C_{5}
\]
The sum becomes:
\[
\sum_{k=0}^{6} {}^{k+5}C_{5} = {}^{5}C_{5} + {}^{6}C_{5} + {}^{7}C_{5} + {}^{8}C_{5} + {}^{9}C_{5} + {}^{10}C_{5} + {}^{11}C_{5}
\]
Step 3: Detailed Calculation
Now we can apply the Hockey-stick identity with $r=5$ and the summation index going from $i=5$ to $n=11$.
\[
\sum_{i=5}^{11} {i \choose 5} = {11+1 \choose 5+1} = {12 \choose 6}
\]
(Note: The first term in our sum is ${}^{5}C_{0} = 1$, which is equal to ${}^{5}C_{5}$. So the transformed sum is correct.)
Now, we compute the value of ${}^{12}C_{6}$:
\[
{}^{12}C_{6} = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!}
\]
\[
= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1}
\]
\[
= \frac{665280}{720} = 924
\]
Step 4: Final Answer
The value of the given sum is 924.
\[
\boxed{924}
\]