Find the total number of 9 digit numbers whic | vedclass

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(A) There are $10$ digits in total: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.$A$ $9$ digit number cannot have $0$ at the first place.Total ways to arrange $9$ d

Vedclass · Accepted Answer

$9 	imes 9!$

Vedclass · Answer

(A) There are $10$ digits in total: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.$A$ $9$ digit number cannot have $0$ at the first place.Total ways to arrange $9$ distinct digits out of $10$ is $^{10}P_9 = \frac{10!}{1!} = 10!$.However,this includes cases where $0$ is at the first place. If $0$ is fixed at the first place,we need to arrange $8$ distinct digits from the remaining $9$ digits in the remaining $8$ positions,which is $^9P_8 = \frac{9!}{1!} = 9!$.Therefore,the number of $9$ digit numbers with all digits different is $^{10}P_9 - ^9P_8 = 10! - 9! = (10 - 1) 	imes 9! = 9 	imes 9!$.

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

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