Correct Answer - A
Key idea Use divisibility test of 11 and consider different situation according to given condition.
Since, the sum of given digits
0 + 1 + 2 + 5 + 7 + 9 = 24
Let the six - digit number be abcdef and to be divisible by 11, so the difference of sum of odd placed digits and sum of even placed digits should be either 0 or a multiple of 11 means `| (a + c + e) - (b + d + f)|` should be either 0 or a multiple of 11.
Hence possible case is a + c + e = 12 = b + d + f (only)
Now, Case I
set {a, c, e} = {0, 5, 7} and set {b, d, f} = {1, 2, 9}
So, number of 6 - digits number `= (2 xx 2!) xx (3!) = 24`
[`because` a can be selected in ways only either 5 or 7].
Case II
Set {a, c, e} = {1, 2, 9} and set {b, d, f} = {0, 5, 7}
So, number of 6 - digits number `= 3! xx 3! = 36`
So, total number of 6 - digits numbers = 24 + 36 = 60
