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consider 5 numbered balls and 5 numbered boxes , how many ways are possible such that exactly 2 balls is not on the ith spot ?

[Ball i  is in the ith box ]

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2!*5c2(1-1+1/2)=10 ?
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i didn't get it can you explain a bit?
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Can anyone tell what is wrong in this? It is exceeding total arrangements of 5! = 120 ways.

1st let us select 2 balls => 5C2 = 10

The 1st ball has 4 positions.

If the 1st ball occupies the position which must not be occupied by 2nd ball, then 2nd ball has 4 positions otherwise if the 1st ball occupies any other position, then 2nd ball has only 3 positions.

The remaining 3 balls can be arranged in 3P3 ways = 3! ways = 6 ways

Thus total ways = 10 * ((1*4)+(3*3)) * 6 = 10*(4+9)*6 = 780 ways
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if the question would have been like this what would be the answer ?

ways such that ball i  is in the ith  box?

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+1

if you arrange rest three balls in 3! ways then  these contains permutation which will have some balls on the correct position which will increase the count on number of balls on the correct place .

I hope u understand  What I wanted to say here?

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Question is incomplete. Can a box remain empty?
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no box remains empty .

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Then it would be 5c2=10.
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10 is correct,isn't it?
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Yeah even I think it should be 10. Put all balls in their respective boxes.

Select any 2 boxes and swap their balls.
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yeah i think so

thanks!