A) 36
B) 60
C) 48
D) 72
Correct Answer: B
Solution :
Sum of given digits 0, 1, 2, 5, 7, 9 is 24. Let the six digit number be abcdef and to be divisible by 11 so \[|(a+c+e)-(b+d+f)|\]is multiple of 11. Hence only possibility is \[a+c+e=12=b+d+f\] Case-I \[\left\{ a,c,e \right\}=\left\{ 9,2,1 \right\}\And \left\{ b,d,f \right\}=\]\[\left\{ 7,5,0 \right\}\] So, Number of numbers \[=3!\times 3!=36\] Case-II \[\left\{ a,c,e \right\}=\left\{ 7,5,0 \right\}\]and \[\left\{ b,d,f \right\}=\left\{ 9,2,1 \right\}\] So, Number of numbers \[2\times 2!\times 3!=24\] Total = 60You need to login to perform this action.
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