(6) Given that $a_{1}, a_{2}, \cdots, a_{2011}$ is a sequence of distinct positive integers, if the order of these 2011 numbers is arbitrarily changed and they are denoted as $b_{1}, b_{2}, \cdots, b_{2011}$, then the value of the number $M=$ $\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right)\left(a_{3}-b_{3}\right) \cdots\left(a_{2011}-b_{2011}\right)$ is $(\quad)$.
(A) must be 0
(B) must be 1
(C) is an odd number
(D) is an even number
6 D Hint: Assume $M$ is odd, then $a_{i}-b_{i}(i=1,2, \cdots, 2011)$ must all be odd, thus the sum of these 2011 odd numbers is also odd; on the other hand, the sum
$$
\begin{array}{c}
\left(a_{1}-b_{1}\right)+\left(a_{2}-b_{2}\right)+\left(a_{3}-b_{3}\right)+\cdots+\left(a_{2011}-b_{2011}\right)= \\
\left(a_{1}+a_{2}+a_{3}+\cdots+a_{2011}\right)-\left(b_{1}+b_{2}+b_{3}+\cdots+b_{2011}\right)=0
\end{array}
$$

is even, a contradiction, so the assumption is false, i.e., $M$ is even.