How many ways are there to choose 6 numbers from the set $\{1,2, \ldots, 49\}$ such that there are at least two consecutive numbers among the 6?
Instead, let's count the number of ways to choose 6 integers (distinct) such that there are no two consecutive integers. Consider the transformation that associates $\left(x_{1}, x_{2}, \ldots, x_{6}\right)$ (assuming the numbers are arranged in ascending order) with $\left(x_{1}, x_{2}-1, x_{3}-2, \ldots, x_{6}-5\right)$. This provides a bijection between the choices of 6 integers from the set $\{1, \ldots, 49\}$ that do not contain consecutive integers and the choices of 6 integers (distinct, but with no additional conditions) from the set $\{1, \ldots, 44\}$. Consequently, there are $\binom{44}{6}$ ways to choose 6 integers from the set $\{1, \ldots, 49\}$ such that there are no consecutive integers. The answer to the question posed is therefore: $\binom{49}{6}-\binom{44}{6}=6924764$.