Find the probablity that at least 2 balls are placed in corresponding number boxes?

Question

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.

My approach.

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.

Does each box only have space for one ball? Or could all the balls go in the same box? — Henry, Nov 17 '12 at 13:01

score 2 · Accepted Answer · answered Nov 17 '12 at 12:45

2

You’re counting many arrangements more than once. As an extreme case, consider the arrangement that has every ball in the right box: it gets counted $\binom62=15$ times, once for every pair of balls.

It’s probably easier to count the arrangements that don’t have at least two balls in the right boxes; here’s a start. Those arrangements are of two types: the derangements of the $6$ balls $-$ the arrangements that have no ball in the right box $-$ and the arrangements that have exactly one ball in the right box. The derangements are a bit messy to count, but the link has a good deal of information to help you with that. Once you know how to count derangements, you can count the arrangements that have exactly one ball in the right box: multiply $6$ ways to choose the ball that’s correctly placed by the number of derangements of the other $5$ balls.

answered Nov 17 '12 at 12:45

Brian M. Scott

616,228

1

A little typical answer.But let me think first. – vikiiii Nov 17 '12 at 12:50
Can you please describe by taking one example. – vikiiii Nov 17 '12 at 13:27
@vikiiii: An example of what? – Brian M. Scott Nov 17 '12 at 13:29
By taking one arrangement having that have no ball in the right box or arrangements that have exactly one ball in the right box. – vikiiii Nov 17 '12 at 13:35
@vikiiii: It’s easy to write down such arrangements: $654321$ has no balls in the right boxes, and $165342$ has exactly one. That’s not what you want to do. I was suggesting that you use the formulas on the Wikipedia page on derangements to count these two types of arrangement. That page is pretty clear, but perhaps this answer to an earlier question would also help. – Brian M. Scott Nov 17 '12 at 13:42
Thanks Brian . This link is also very useful http://mathworld.wolfram.com/Derangement.html – vikiiii Nov 17 '12 at 18:19
Is there is any easier way to divide any number by e ? – vikiiii Nov 19 '12 at 12:52
@vikiiii: Easier than what? – Brian M. Scott Nov 19 '12 at 17:30
e has the value 2.7182 . Now suppose divide 6! by e will be very complicated .So i am thinking that if e can be presented as a numerator/denominator which can be easy to solve. – vikiiii Nov 20 '12 at 05:49
@vikiiii: No, it can’t: $e$ is irrational. (It’s even transcendental.) – Brian M. Scott Nov 20 '12 at 06:07

Find the probablity that at least 2 balls are placed in corresponding number boxes?

1 Answers1