1982 AMC 12 Problem 11

First count the allowed ordered pairs of first and last digits

Once those digits are fixed, choose two distinct middle digits from the remaining eight

For leading digits $1,\ldots,9,$ the number of possible last digits differing by $2$ is $1+1+2+2+2$ plus $2+2+1+1,$ or $14.$ In addition, leading digit $2$ may end in $0,$ giving $15$ endpoint pairs. The middle digits can then be chosen in $8\cdot7$ ordered ways. Thus the count is $15\cdot56=840.$

Therefore, the correct answer is C.