An eight-digit number divisible by 9 is to be | vedclass

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(D) number is divisible by $9$ if the sum of its digits is divisible by $9$.The sum of all digits from $0$ to $9$ is $0+1+2+3+4+5+6+7+8+9 = 45$.To for

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$36$

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(D) number is divisible by $9$ if the sum of its digits is divisible by $9$.The sum of all digits from $0$ to $9$ is $0+1+2+3+4+5+6+7+8+9 = 45$.To form an $8$-digit number,we must exclude two digits such that the sum of the remaining $8$ digits is a multiple of $9$. Since the total sum is $45$,the sum of the two excluded digits must be $0+9=9$,$1+8=9$,$2+7=9$,$3+6=9$,or $4+5=9$.						Excluded Digits			Number of $8$-digit numbers							$0$ and $9$			$8! = 8 	imes 7!$							$1$ and $8$			$8! - 7! = 7 	imes 7!$							$2$ and $7$			$8! - 7! = 7 	imes 7!$							$3$ and $6$			$8! - 7! = 7 	imes 7!$							$4$ and $5$			$8! - 7! = 7 	imes 7!$			Total ways $= 8 	imes 7! + 4 	imes (7 	imes 7!) = 8 	imes 7! + 28 	imes 7! = 36 	imes 7!$.

An eight-digit number divisible by $9$ is to be formed using digits from $0$ to $9$ without repeating the digits. The number of ways in which this can be done is: (in $(7!)$)

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