For any two numbers $n$ and $k$ it is always true that $$\binom{n}{k} = \binom{n}{n-k}.$$ We can see that from the formula for $\displaystyle \binom{n}{k}$: $$\binom{n}{k} = \frac{n!}{k!(n-k)!} \quad\text{and}\quad \binom{n}{n-k} = \frac{n!}{(n-k)!(n-(n-k))!} = \frac{n!}{(n-k)!k!}.$$

In your example they are indeed equal: $$\binom{13}{6} = \frac{13!}{6!7!} = \frac{13!}{7!6!} = \binom{13}{7}.$$ Note this is true because $6+7 = 13$.