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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.

  • 0
    Does each box only have space for one ball? Or could all the balls go in the same box?2012-11-17
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    Space for one ball.2012-11-17

1 Answers 1

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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.

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    A little typical answer.But let me think first.2012-11-17
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    Can you please describe by taking one example.2012-11-17
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    @vikiiii: An example of what?2012-11-17
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    By taking one arrangement having that have no ball in the right box or arrangements that have exactly one ball in the right box.2012-11-17
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    @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](http://math.stackexchange.com/a/83472/12042) to an earlier question would also help.2012-11-17
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    Thanks Brian . This link is also very useful http://mathworld.wolfram.com/Derangement.html2012-11-17
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    Is there is any easier way to divide any number by `e` ?2012-11-19
  • 0
    @vikiiii: Easier than what?2012-11-19
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    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.2012-11-20
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    @vikiiii: No, it can’t: $e$ is irrational. (It’s even [transcendental](http://en.wikipedia.org/wiki/Transcendental_number).)2012-11-20