AMC 8 Daily Practice Round 11

Complete problem set with solutions and individual problem pages

Problem 16 Medium

How many four-digit numbers satisfy the condition that the digit in the units place is greater than the digit in the thousands place, the digit in the thousands place is greater than the digit in the hundreds place, and the digit in the hundreds place is greater than the digit in the tens place?

  • A.

    5040

  • B.

    3024

  • C.

    840

  • D.

    210

  • E.

    105

Answer:D

Since the digits of the four-digit number must follow a strict order (units digit > thousands digit > hundreds digit > tens digit), each selection of 4 distinct digits from 0 to 9 corresponds to exactly one valid number. This is because the order of the digits is uniquely determined by their sizes. Note that 0 can be included in the selection, as it will occupy the smallest position (tens place) if chosen, satisfying the requirement that the leading digit (thousands place) cannot be 0. The total number of ways to choose 4 digits from 0 to 9 is:

\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210.

Thus, there are 210 such four-digit numbers.