6. From $1,2, \cdots, 49$, choose six different numbers, among which at least two are adjacent. The total number of ways to choose is ( ).
(A) $C_{49}^{6}$
(B) $C_{44}^{6}$
(C) $C_{49}^{6}-C_{44}^{5}$
(D) $C_{19}^{6}-C_{44}^{6}$
6. D Let $a_{1}, a_{2}, \cdots, a_{6}$ be six different numbers taken from $1,2, \cdots, 49$. Without loss of generality, assume $a_{1}<a_{2}<\cdots<a_{6}$. Clearly, $a_{1} \leqslant a_{2}-1 \leqslant a_{3}-2 \leqslant a_{4}-3 \leqslant a_{5}-4 \leqslant a_{6}-5$, and the necessary and sufficient condition for $a_{1}, a_{2}-1, a_{3}-2, a_{4}-3, a_{5}-4, a_{6}-5$ to be distinct is that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ do not contain adjacent numbers. Let the six-tuple $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right)$ correspond to $\left(a_{1}, a_{2}-1, a_{3}-2, a_{4}-3, a_{5}-4, a_{6}-5\right)$. Then a one-to-one correspondence is established between the set of six-tuples consisting of six different and non-adjacent numbers taken from 1 to 49 and the set of six-tuples consisting of six different numbers taken from 1 to 44. Therefore, the number of six-tuples in both sets is equal to $C_{44}^{6}$. The number of six-tuples consisting of six different numbers taken from 1 to 49 is $C_{49}^{6}$. Thus, the number of six-tuples with adjacent numbers is $C_{49}^{6}-C_{44}^{6}$.