The number of integers between 10 and 10,000 | vedclass

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(C) We need to find the number of integers $n$ such that $10 < n < 10,000$ and the digits of $n$ are in strictly increasing order.This means we are lo

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$246$

Vedclass · Answer

(C) We need to find the number of integers $n$ such that $10 This means we are looking for integers with $2, 3,$ or $4$ digits.For any set of $k$ distinct digits chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,there is exactly one way to arrange them in strictly increasing order.Note that $0$ cannot be included because if $0$ is present,it would have to be the first digit to maintain the increasing order,which is not allowed for integers between $10$ and $10,000$.$1$. For $2$-digit integers: We choose $2$ digits from $9$ available digits. The number of ways is $\binom{9}{2} = \frac{9 	imes 8}{2} = 36$.$2$. For $3$-digit integers: We choose $3$ digits from $9$ available digits. The number of ways is $\binom{9}{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.$3$. For $4$-digit integers: We choose $4$ digits from $9$ available digits. The number of ways is $\binom{9}{4} = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.Total number of such integers $= 36 + 84 + 126 = 246$.

The number of integers between $10$ and $10,000$ such that in every integer every digit is greater than its immediate preceding digit,is

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