How many numbers can be formed using digits 1 2 3 4 5 6 7 8 9 such that they are in increasing order(e.g. 12345, 578)? Help me in how to solve this problem please. I will be grateful to you :)
Consider the string $123456789$. Observe that each of the $9$ digits in the string can safely be either included or not included, since failing to include a digit will not violate the increasing order of the digits. Hence, there are $2^9$ possible strings that contain ($0$ or $1$ or $2$ or ... or $9$) digits. However, we likely want to omit the case where there are $0$ digits, which leaves us with a final answer of:
$$
2^9 - 1 = 511
$$
numbers with digits in increasing order where they are in a.p. With $r=1 ...$ If it's a $5$ digit no and no starts with $5$ , then the number would be $56789$ . .. As I explained the question properly , now m pretty sure , u urself would be able to solve d problem , one digit numbers would not be considered .. $2$ digit no.- $8$ $(12,23,34,45,56,67,78,89)$
$3$ digit no.- $7(123,234,345,456,567,678,789)$,
$4$ digit no.- $6$, $5$ digit no -$5$, $6$ digit no -$4$, $7$ digit no-$3$, $8$ digit no-$2$, $9$ digit no -$1(123456789)$ ,... So total $=1+2+3+•••+8=36 $