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 and no digit is repeated, is :

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 $\{a, c, e\}=\{9,2,1\} \&\{b, d, f\}=$ $\{7,5,0\}$

$\{7,5,0\}$

So, Number of numbers $=3 ! \times 3 !=36$

Case-II $\{a, c, e\}=\{7,5,0\}$ and $\{b, d, f\}=\{9,2,1\}$

So, Number of numbers $2 \times 2 ! \times 3 !=24$

Total $=60$