2 votes 2 votes 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? Quantitative Aptitude cat2018-1 quantitative-aptitude permutation-combination numerical-answer + – go_editor asked Mar 19, 2020 • retagged Mar 16, 2022 by Lakshman Bhaiya go_editor 13.9k points 722 views answer comment Share See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Given that, the digits $:1,2,3,4,5,6,7,8,9 $ We know that, the number of ways to pick $k$ unordered elements from an $n$ element set $ = \;^{n}C_{k} = \frac{n!}{k!(n-k)!} $ After selecting the number from the given digits, there is only one way to arrange it. So, the total number of ways $ = \;^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} \quad \longrightarrow (1)$ We know that, $^{n} C_{0} + \;^{n} C_{1} + \;^{n} C_{2} + \dots + \;^{9} C_{n} = 2^{n}$ Here, $^{9}C_{0} + \;^{9}C_{1} + \;^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} = 2^{9}$ $\Rightarrow \; ^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} = 2^{9} – \;^{9}C_{0} - \;^{9}C_{1}$ From the equation $(1),$ we get $\therefore$ The total number of ways $ = 2^{9} – \;^{9}C_{0} – \;^{9}C_{1}$ $\quad = 512 – \frac{9!}{0! \cdot 9!} – \frac{9!}{1! \cdot 8!} $ $\quad = 512 – 1 – \frac{9 \times 8!}{1! \cdot 8!} $ $\quad = 512 – 1 – 9 $ $\quad = 512 – 10 $ $\quad = 502 \; \text{ways}.$ Correct Answer $: 502$ $ \textbf{PS}:\text{Important Properties:}$ $n! = n(n-1)(n-2) \dots 1 = n(n-1)!$ $0! = 1 $ $1! = 1 $ $ ^{n}C_{n} = \frac{n!}{n! \; 0!} = 1 $ $ ^{n}C_{0} = \frac{n!}{0! \; n!} $ $ ^{n}C_{1} = \frac{n!}{1!(n-1)!} = \frac{n(n-1)!}{1!(n-1)!} = n $ Anjana5051 answered Sep 17, 2021 • edited Sep 18, 2021 by Anjana5051 Anjana5051 11.7k points comment Share See all 0 reply Please log in or register to add a comment.