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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]

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Let us consider the case of 2-digit numbers

2 numbers can be chosen from the given set by ^{9}C_{2} 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 ^{9}C_{2}+^{9}C_{3}+......+^{9}C_{8}+^{9}C_{9} ways

We know, ^{n}C_{0}+^{n}C_{1}+.....+^{n}C_{n-1}+^{n}C_{n} = 2^{n}

So, the total number of ways = 2^{9} – ^{9}C_{0} – ^{9}C_{1}

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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