Distinct Digits Order Permutation

We need to form 4-digit numbers with distinct digits from $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that 7 appears to the left of 3.

Case 1: Both 7 and 3 are present in the 4-digit number. First, select 2 additional digits from the remaining 8 digits (excluding 7 and 3): $\binom{8}{2} = 28$ ways. Now we have 4 digits: 7, 3, and two others. Number of arrangements of 4 distinct digits such that 7 comes before 3 is $\frac{4!}{2} = 12$. Total permutations without zero condition = $28 \times 12 = 336$. Now subtract cases where 0 is at the thousand's place: If 0 is at thousands place, we select 1 more digit from remaining 7 digits: $\binom{7}{1} = 7$ ways. The remaining 3 places are filled by 7, 3, and the selected digit. Arrangements with 7 before 3 in the 3 remaining positions $= \frac{3!}{2} = 3$. Invalid numbers starting with 0 $= 7 \times 3 = 21$. Valid numbers in Case 1 $= 336 - 21 = 315$.

Since 7 coming before 3 requires both 7 and 3 to be present, there are no other cases. Total count = 315.