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

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(C) 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 $45$. We need to choose $8$ digi

Vedclass · Accepted Answer

$36 	imes 7!$

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(C) 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 $45$. We need to choose $8$ digits such that their sum is divisible by $9$. Let the two excluded digits be $x$ and $y$. Then $45 - (x + y)$ must be divisible by $9$, which implies $(x + y)$ must be $0$ or $9$ or $18$. Since $x, y \in \{0, 1, \dots, 9\}$ and $x 
eq y$, the possible pairs $(x, y)$ are: $(i)$ $x+y=0$: $(0, 0)$ (not possible as digits are distinct). (ii) $x+y=9$: $(0, 9), (1, 8), (2, 7), (3, 6), (4, 5)$ ($5$ pairs). (iii) $x+y=18$: $(9, 9)$ (not possible as digits are distinct). Case $1$: Excluded pair is $(1, 8), (2, 7), (3, 6), (4, 5)$. For each pair, we have $8!$ permutations. Total $= 4 	imes 8! = 4 	imes 8 	imes 7! = 32 	imes 7!$. Case $2$: Excluded pair is $(0, 9)$. The first digit cannot be $0$. Total permutations $= 8! - 7! = 7! 	imes (8 - 1) = 7 	imes 7!$. Total ways $= 32 	imes 7! + 7 	imes 7! = 39 	imes 7!$. Wait, re-evaluating: The sum of all $10$ digits is $45$. We choose $8$ digits. Sum of $8$ digits $= 45 - (x+y)$. For this to be divisible by $9$, $x+y$ must be $0, 9, 18$. Pairs $(x, y)$ with $x+y=9$: $(0,9), (1,8), (2,7), (3,6), (4,5)$. If $(x,y) = (0,9)$, digits are $\{1,2,3,4,5,6,7,8\}$. Number of ways $= 8! = 40320$. If $(x,y) \in \{(1,8), (2,7), (3,6), (4,5)\}$, digits include $0$. Number of ways $= 7 	imes 7! = 35280$. Total $= 8! + 4 	imes (7 	imes 7!) = 8 	imes 7! + 28 	imes 7! = 36 	imes 7!$.

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

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