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43 votes

In how many ways can we distribute $5$ distinct balls, $B_1, B_2, \ldots, B_5$ in $5$ distinct cells, $C_1, C_2, \ldots, C_5$ such that Ball $B_i$ is not in cell $C_i$, $\forall i= 1,2,\ldots 5$  and each cell contains exactly one ball?

  1. $44$
  2. $96$
  3. $120$
  4. $3125$
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11 Answers

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The number of derangements of distinct "n" elements is = ! n

 

!n = (n-1)×(!(n-1) + !(n-2))  where  [ !1 = 0, !2 = 1]

 

!3 = (3-1)×(!(3-1) + !(3-2) ) = 2(1+0) = 2

 

!4 = (4-1)×(!(4-1) +! (4-2)) =3(2+1) = 9

 

!5 =(5-1)×(!(5-1) + !(5-1)) = 4(9 +2) = 44

 

∴Options is the correct answer.

useful link-https://www.youtube.com/watch?v=ZT8zBvCr7L8

 

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This is a direct derangement problem where n = 5

No. Of derangement = 5!*(1/2!-1/3!+1/4!-1/5!)

=44
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Although there are many methods out there,  but this one was cool so i posted it.

But we are asked the Derangements that they must not go in their original box.

So that would be

 

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