$32 \cdot 31$ is a four-digit integer between 1000 and 9999, composed of four different digits, the absolute difference between the first and the last digit is 2. The number of such integers is
(A) 672.
(B) 784.
(C) 840.
(D) 896.
(E) 1008.
(33rd American High School Mathematics Examination, 1982)
[Solution] First, consider how many pairs of numbers in the set $\{0,1, \cdots, 9\}$ have a difference of $\pm 2$.
Obviously, there are 16 pairs: $(0,2),(2,0),(1,3),(3,1),(2,4),(4,2), \cdots,(7$, $9),(9,7)$. Except for $(0,2)$, all of them can be used to form the required number.

Since the digits of the four-digit number to be formed must all be different, the possible scenarios for the middle two digits are a total of $8 \cdot 7=56$.
Therefore, the total number of numbers that meet the requirements of the problem is $15 \cdot 56=840$.
Thus, the answer is $(C)$.