Problem:
How many integers between and have four distinct digits arranged in increasing order? (For example, is one integer.)
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Since the integers are between and , we know the thousands digit must be and the hundreds digit must be between and . Since the digits are increasing, the second digit must be greater than , so it can only be . This means the tens and units digits are different digits, each greater than or equal to .
Suppose we choose distinct digits, each greater than or equal to . There are digits for the first choice and digits for the second choice. This means there are combinations. However, we ignore exactly half of these combinations, as each combination has an equal likelihood of being ascending or descending. This leaves combinations.
Thus, the correct answer is .
Answer: .
The problems on this page are the property of the MAA's American Mathematics Competitions