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(B) Let the $6$-digit number be $abcdef$. For the number to be divisible by $11$,the condition $|(a+c+e) - (b+d+f)|$ must be a multiple of $11$.Since

Vedclass · Accepted Answer

$60$

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(B) Let the $6$-digit number be $abcdef$. For the number to be divisible by $11$,the condition $|(a+c+e) - (b+d+f)|$ must be a multiple of $11$.Since the sum of all digits is $0+1+2+5+7+9 = 24$,let $S_1 = a+c+e$ and $S_2 = b+d+f$. Then $S_1 + S_2 = 24$ and $S_1 - S_2 = 11k$.For $k=0$,$S_1 = S_2 = 12$.Possible sets for ${a, c, e}$ and ${b, d, f}$ such that their sum is $12$:Case $I$: ${a, c, e} = {9, 2, 1}$ and ${b, d, f} = {7, 5, 0}$.Number of ways to arrange ${a, c, e}$ is $3! = 6$. Number of ways to arrange ${b, d, f}$ is $3! = 6$. Total $= 6 	imes 6 = 36$.Case $II$: ${a, c, e} = {7, 5, 0}$ and ${b, d, f} = {9, 2, 1}$.Here $a 
eq 0$. If $a$ is chosen from ${7, 5}$,there are $2$ choices. The remaining $2$ positions for ${c, e}$ can be filled in $2!$ ways. The set ${b, d, f}$ can be arranged in $3! = 6$ ways.Total $= 2 	imes 2! 	imes 6 = 24$.Total numbers $= 36 + 24 = 60$.

The number of $6$-digit numbers that can be formed using the digits $0, 1, 2, 5, 7,$ and $9$ which are divisible by $11$,where no digit is repeated,is:

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