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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. $3125$
in Combinatory
edited by
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34

Option elimination for this problem only
Total arrangements possible = 5! = 120

1 combination when all balls are at their respective places
so remaining arrangements = 120 - 1 = 119 
this eliminates Options C & D

when B1 is placed at C1
no. of arrangements = 1 * 4 ! = 24
so remaining arrangements = 119 - 24 = 95 
this eliminates Options B 

Use this method when you dont know how to solve the ques
and their is large difference in the given options like above options

7

Before checking answer first read Derangement concept.

11 Answers

0 votes

Use the following formula to calculate the number of derangement :

$ \left \lfloor \frac{n!}{e} +\frac{1}{2}\right \rfloor \quad $ 

 $ \left \lfloor \frac{n!}{e} +\frac{1}{2}\right \rfloor \quad $ 

=$ \left \lfloor \frac{5!}{e} +\frac{1}{2}\right \rfloor \quad $

=$ \left \lfloor \frac{120}{2.71828} +\frac{1}{2}\right \rfloor \quad $

=$44$  

0 votes

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

 

0 votes
This is a direct derangement problem where n = 5

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

=44
Answer:

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