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

The correct answer is 315.

We need to count all 4-digit numbers where:

(i) The number is greater than 1000 (i.e., the leading digit is not 0)
(ii) All four digits are distinct
(iii) The digit 7 appears before the digit 3 (both must be present)

Since 7 must come before 3, both digits 7 and 3 must be present in the number. We then choose 2 more distinct digits from the remaining 8 digits: {0, 1, 2, 4, 5, 6, 8, 9}.

Step 1: Choose the remaining 2 digits

From the 8 available digits, choose 2:

$C_{}^{}(8,2)=28$

We split these 28 groups into two cases based on whether 0 is included or not.

Step 2 — Case A: 0 is NOT one of the two extra digits

We choose 2 digits from {1, 2, 4, 5, 6, 8, 9} (7 digits, no zero):

$(7,2)=21\text{groups}$

Each group gives us a set of 4 digits: {3, 7, digit₁, digit₂}. Since none of them is 0, the leading digit restriction is automatically satisfied.

Total arrangements of 4 distinct non-zero-leading digits = 4! = 24.

Among all arrangements, by symmetry, exactly half have 7 appearing before 3:

$\frac{24}{2}=12\text{valid arrangements per group}$

$\text{Case A total}=21\times 12=252$

Step 3 — Case B: 0 IS one of the two extra digits

We choose 1 more digit from {1, 2, 4, 5, 6, 8, 9}:

$(7,1)=7\text{groups}$

Each group gives us a set of 4 digits: {0, 3, 7, digit_x}. Now we must exclude arrangements where 0 is the leading digit.

Total arrangements with 7 before 3 (ignoring leading-zero rule):

$\frac{4!}{2}=12$

Invalid arrangements (0 is the leading digit AND 7 before 3):
Fix 0 in the first position. The remaining 3 positions hold {3, 7, digit_x}. We need 7 to appear before 3 among these 3 positions. The number of ways to arrange 3 items such that 7 comes before 3:

$\frac{3!}{2}=3\text{invalid arrangements per group}$

Valid arrangements per group = 12 − 3 = 9

$\text{Case B total}=7\times 9=63$

Step 4: Add both cases

$\text{Total}=\text{Case A}+\text{Case B}=252+63=$315

Therefore, the total number of valid 4-digit numbers greater than 1000, with all distinct digits, and with 7 coming before 3, is 315.