How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?
Explanation:
Out of 4 digits we already have 7 and 3. Now we have to select 2 more digits.
Case 1: zero is not selected.
We can select 2 more digits out of 7 in 7C2 = 21 ways.
These 4 digits can be arranged in 4! = 24 ways
∴ Total number of ways = 21 × 24 = 504 ways.
Now, in half of these numbers 7 would be before 3 and in other half 3 would be before 7.
∴ 7 comes before 3 in 504/2 = 252 ways.
Case 2: zero is selected.
We can select 1 more digit out of 7 in 7C1 = 7 ways.
These 4 digits can be arranged in 3 × 3! = 18 ways
∴ Total number of ways = 7 × 18 = 126 ways.
∴ 7 comes before 3 in 126/2 = 63 ways.
∴ Total numbers in which 7 comes before 3 = 252 + 63 = 315.
Hence, 315.
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