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 $45$. To form an $8$-digit numbe

Vedclass · Accepted Answer

$36$

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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 $45$. 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 $4+5=9$.Case $1$: Exclude ${0, 9}$. The remaining $8$ digits can be arranged in $8!$ ways. However,the first digit cannot be $0$. Since $0$ is already excluded,all $8!$ arrangements are valid. Total $= 8! = 8 	imes 7!$.Case $2, 3, 4, 5$: Exclude ${1, 8}, {2, 7}, {3, 6},$ or ${4, 5}$. For each pair,the number of $8$-digit arrangements is $8! - 7!$ (total permutations minus those starting with $0$). This equals $7 	imes 7!$.Total ways $= (8 	imes 7!) + 4 	imes (7 	imes 7!) = 8(7!) + 28(7!) = 36(7!)$.Excluded DigitsNumber of $8$-digit numbers${0, 9}$$8 	imes 7!$${1, 8}$$7 	imes 7!$${2, 7}$$7 	imes 7!$${3, 6}$$7 	imes 7!$${4, 5}$$7 	imes 7!$

$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y + 5 = 0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x = 1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x + 8y = 27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x = 9$

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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