Question:medium

How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3

Updated On: Jan 15, 2026
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Solution and Explanation

Assumptions:

  • Allowed digits: 0-9 (10 total).
  • All digits within a number must be unique.
  • A 4-digit number cannot have 0 as its first digit.

Methodology:

The problem is divided into two distinct scenarios:

Scenario 1: The digit 7 occupies the thousands place.

  • The thousands digit is fixed as 7.
  • Three additional digits must be selected from the remaining 9 available digits (excluding 7).
  • One of these three selected digits must be 3.
  • The digit 3 can be placed in any of the 3 remaining positions, offering 3 possibilities.
  • The remaining 2 digits are chosen from the 8 digits that are neither 7 nor 3:

The total for this scenario is calculated as: \[ \text{Scenario 1} = 3 \times {}^8P_2 = 3 \times 8 \times 7 = 168 \]

Scenario 2: The digit 7 is NOT in the thousands place.

  • The thousands digit can be any digit except 0, 3, or 7. This leaves 7 choices.
  • Both 7 and 3 must be present in the remaining 3 positions.
  • Two positions are selected for 7 and 3, yielding \({}^3P_2 = 3 \times 2 = 6\) permutations. However, since we are placing the specific digits 7 and 3, there are 3 unique arrangements for placing them.
  • The final digit for the 4-digit number is selected from the 7 remaining digits (excluding 0, 3, 7, and the digit used in the thousands place), offering 7 possibilities.

The total for this scenario is calculated as: \[ \text{Scenario 2} = 7 \times 3 \times 7 = 147 \]

Total 4-digit numbers meeting the criteria:

\[ \text{Total} = 168 + 147 = \boxed{315} \]

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