Find the total number of 9 digit numbers whic | vedclass

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(A) There are $10$ digits in total,namely $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.To form a $9$ digit number with all distinct digits,the first digit cannot be

Vedclass · Accepted Answer

$9 	imes 9!$

Vedclass · Answer

(A) There are $10$ digits in total,namely $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.To form a $9$ digit number with all distinct digits,the first digit cannot be $0$.For the first position,there are $9$ choices ($1$ to $9$).For the remaining $8$ positions,we need to choose $8$ digits from the remaining $9$ digits (including $0$),which can be arranged in $P(9, 8)$ ways.Total numbers = $9 	imes P(9, 8) = 9 	imes \frac{9!}{(9-8)!} = 9 	imes 9!$.Alternatively,total $9$ digit numbers with distinct digits = (Total permutations of $9$ digits out of $10$) - (Permutations starting with $0$):$= P(10, 9) - P(9, 8) = \frac{10!}{1!} - \frac{9!}{1!} = 10! - 9! = (10 - 1) 	imes 9! = 9 	imes 9!$.

Find the total number of $9$ digit numbers which have all the digits different.

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