How many numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?
Explanation:
As the digits appear in ascending order, there is only one way in which selected group of integers can be arranged.
Thus for two digit number, we need to select two digits.
This can be done in 9C2 = 36 ways
As nCm =9C(n−m) ; 9C2 = 9C7 = Number of ways in which number can have 2 or 7 digits = 9C2 = 9C7 = 36 ways
Number of ways in which number can have 3 or 6 digits = 9C3 = 84 ways
Number of ways in which number can have 4 or 5 digits = 9C4 = 126 ways
Number of ways in which number can have 8 digits = 9C8 = 8 ways
Number of ways in which number can have 9 digits = 1 ways
1 + 8 + 2(36) + 2(84) + 2(126) = 502 numbers have 2 or more distinct digits such that the digits appear in ascending order.
Hence, 502.
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