There are six balls and six boxes numbered 1 to 6.
So I have to find the probablity that at least 2 balls are placed in corresponding number boxes.
Out of 6 balls , select 2 which will be placed in right boxes.
So it will be $\binom62$.
Now remaining 4 balls can be arranged in any ways so it will be .
So total will be $\binom62 \times 4!$.
And the denominator will be $6!$.
So answer comes to be $0.5$ .
But the answer is $1/60$ .
Where am i wrong ?
Because i think that i have done it a right way.
Thanks in advance.