Question

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

Answer 1

Assumptions:

• Only digits from 0 to 9 can be used (i.e., total 10 digits).
• Each digit in the number must be distinct.
• 4-digit number cannot start with 0.

Approach:

We break the problem into two mutually exclusive cases:

Case 1: 7 is in the thousand's place

• The thousand's digit is 7, so it's fixed.
• We now need to choose 3 more digits from the remaining 9 digits (excluding 7).
• Among those 3, one digit must be 3.
• We place 3 in one of the 3 remaining positions ⇒ 3 ways
• The remaining 2 digits are chosen from the 8 remaining digits (excluding 7 and 3):

So the total for this case is: \[ \text{Case 1} = 3 \times {}^8P_2 = 3 \times 8 \times 7 = 168 \]

Case 2: 7 is NOT in the thousand's place

• Thousand's digit ≠ 0, 3, or 7 ⇒ choose from the other 7 digits ⇒ 7 ways
• In the remaining 3 positions, both 7 and 3 must appear.
• We choose 2 positions for 7 and 3 ⇒ \({}^3P_2 = 3 \times 2 = 6\) ways, but since one is 7 and one is 3, we just count 3 ways to place 7 and 3.
• The 4th position (to complete 4-digit number) is chosen from remaining 7 digits (excluding 0, 3, 7 and thousand digit) ⇒ 7 ways

So the total for this case is: \[ \text{Case 2} = 7 \times 3 \times 7 = 147 \]

Total 4-digit numbers satisfying the condition:

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

Answer 2

Approach:

We solve this by considering two distinct cases based on the position of digit 7.

Case 1: 7 is at the leftmost (thousands) place

• The thousands digit is fixed as 7.
• Now, digit 3 must be in one of the remaining 3 positions ⇒ 3 ways
• We now need to fill the remaining 2 digits using any of the digits from {0, 1, 2, 4, 5, 6, 8, 9} (8 choices), excluding 3 and 7 ⇒ choose 2 from 8 in 8 × 7 ways

So, total numbers in this case: \[ 3 \times 8 \times 7 = 168 \]

Case 2: 7 is not at the leftmost place

• The thousands digit cannot be 0, 3, or 7 ⇒ choose from the other 7 digits ⇒ 7 options
• Now, 3 and 7 must be placed in the remaining 3 positions ⇒ choose 2 positions out of 3 for them ⇒ 3 options
• The remaining digit (to make total 4 digits) can be chosen from remaining digits (excluding 3, 7, and the digit already placed at the first position) ⇒ 7 options

So, total numbers in this case: \[ 7 \times 3 \times 7 = 147 \]

Final Answer:

Add both cases: \[ 168 + 147 = \boxed{315} \]