How many 4-digit numbers can be formed such that 2 always comes before 8, if repetition of digits is not allowed?
Explanation:
2 and 8 should definitely be included, hence we need to select two more digits out of 3, 4 and 7. This can be done in 3C2 = 3 ways. Number of ways of arranging any selection of 4 digits = 4! = 24
∴ Total number of 4-digit numbers without repetition = 3 × 4! = 72.
Out of these 72 numbers, half of them will have 2 before 8 and the other half will have 8 before 2.
∴ Number of 4-digit numbers that can be formed such that 2 always comes before 8 = 72/2 = 36.
Hence, 36.
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