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Sum of the given digit is 0+1+2+5+7+9 = 24
6 digit number let abcdef
abcdef is divisible by 11 if
Only one possible is there
a+c+e=b+d+f=12
Case 1: {a,c,e}={7,5,0}
{b,d,f}={9,2,1}
So, 2 x 2! x 3! = 24
Case 2: {a,c,e}={9 , 2, 1}
{b,d,f}={7, 5, 0}
So, 3! x 3! = 36
Total = 24 + 36 = 60
So, correct option is (2).
Option 1)
72
Option 2)
60
Option 3)
48
Option 4)
36