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

We need to find 4-digit numbers where each number is greater than 1000 (so first digit can't be 0), all four digits are distinct (no repeated digits), and the digit 7 appears before the digit 3 in the number.

Key Insight: Since 7 must come before 3, we need to consider all possible positions where 7 can be placed, with 3 coming after it.

---

We'll use case-by-case analysis based on where 7 is positioned, ensuring 3 always comes after it.

When 7 is in the first position: $7 \_ \_ \_$

The digit 3 can be placed in any of the remaining three positions (2nd, 3rd, or 4th).

7 is 1st, 3 is 2nd → $7 3 \_ \_$

First position: 7 (fixed) → 1 way

Second position: 3 (fixed) → 1 way

Third position: Any digit except 7,3 → 8 ways

Fourth position: Any remaining digit → 7 ways

Total: $1 \times 1 \times 8 \times 7 = 56$

7 is 1st, 3 is 3rd → $7 \_ 3 \_$

First position: 7 (fixed) → 1 way

Second position: Any digit except 7,3 → 8 ways

Third position: 3 (fixed) → 1 way

Fourth position: Any remaining digit → 7 ways

Total: $1 \times 8 \times 1 \times 7 = 56$

7 is 1st, 3 is 4th → $7 \_ \_ 3$

First position: 7 (fixed) → 1 way

Second position: Any digit except 7,3 → 8 ways

Third position: Any remaining digit → 7 ways

Fourth position: 3 (fixed) → 1 way

Total: $1 \times 8 \times 7 \times 1 = 56$

Case 1 Total: $56 + 56 + 56 = 168$

---

When 3 is in the last position: $\_ \_ \_ 3$

The digit 7 must be in one of the first three positions (since 7 must come before 3).

7 is 2nd, 3 is 4th → $\_ 7 \_ 3$

First position: Any digit except 0,7,3 → 7 ways

Second position: 7 (fixed) → 1 way

Third position: Any remaining digit → 7 ways

Fourth position: 3 (fixed) → 1 way

Total: $7 \times 1 \times 7 \times 1 = 49$

7 is 3rd, 3 is 4th → $\_ \_ 7 3$

First position: Any digit except 0,7,3 → 7 ways

Second position: Any remaining digit → 7 ways

Third position: 7 (fixed) → 1 way

Fourth position: 3 (fixed) → 1 way

Total: $7 \times 7 \times 1 \times 1 = 49$

Case 2 Total: $49 + 49 = 98$

---

When both 7 and 3 are in positions 2 and 3: $\_ 7 3 \_$

First position: Any digit except 0,7,3 → 7 ways

Second position: 7 (fixed) → 1 way

Third position: 3 (fixed) → 1 way

Fourth position: Any remaining digit → 7 ways

Total: $7 \times 1 \times 1 \times 7 = 49$

---

Adding all cases together:

Total = $168 + 98 + 49 = 315$

Important Note: We used the constraint that the first digit cannot be 0 (since we need 4-digit numbers greater than 1000). This is why when 7 is not in the first position, we have only 7 choices for the first digit instead of 8.