Track / Stage 4 / 260 of 340 #520 of 2000

Problem 520

AMC 12 late, AIME early

Combinatorics Difficulty 5.0 Find the answer

4. Among the integers between 1000 and 9999, the number of four-digit integers with all different digits, where the absolute difference between the first and last digit is 2, is
A. 672
B. 784
C. 840
D. 896

Your answer

Official solution

4. C 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 these can be used to form the required number.

Since the requirement is that the digits of the four-digit number must all be different, the possible scenarios for the middle two digits are $8 \cdot 7=$ 56 in total.
Therefore, the total number of numbers that meet the requirement is $15 \cdot 56=840$ .
Hence, the answer is C.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.