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Question 14 : How many numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 so that in every such number, each digit is used at most once and the digits appear in the ascending order? [TITA]
Let us consider the case of 2-digit numbers
2 numbers can be chosen from the given set by 9C2 ways. There is only one way of arranging them in ascending order.
Similarly, the remaining numbers can be chosen and arranged in ascending order by 9C2+9C3+......+9C8+9C9 ways
We know, nC0+nC1+.....+nCn-1+nCn = 2n
So, the total number of ways = 29 – 9C0 – 9C1
512-10 = 502
The question is "How many numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 so that in every such number, each digit is used at most once and the digits appear in the ascending order?"
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