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Using the six digits 1,3,5,6,7,9 once at a time how many 6 digit numbers can be built which are divisible by 11?

We know the divisibility criteria for 11. That is let $abcdef$ be the number. where $abcdef$ represents the digits. Then $abcdef$ is divisible by 11 if $(a+c+e)-(b+d+f)$ is divisible by 11.
We see,
$(9+7+5)-(1+3+6)=11$ So this is a valid configuration of the number. And we can change the positions of the numbers comprised in the brackets. so the total number of permutations possible for the number is $3! \times 3!=36$.
Again we can do another thing,$(1+3+6)-(9+7+5)=-11$. This will give us $36$ more numbers. So the total number of numbers we have is $72$. If we exchange any one number in the sets $\{9,7,5\}$ and $\{1,3,6\}$ the difference will not be a multiple of $11$. Again if we change $2$ of the numbers, the difference cannot be a multiple of $11$.

Thank you so much. It 's seeming very easy to me after you solved the problem. Thanks bhai.