GATE IT 2004 | Question: 35

Question

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?

• A. $44$
• B. $96$
• C. $120$
• D. $3125$

Comments

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

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Before checking answer first read Derangement concept.

Answer 1

Number of derangements of n-objects = $n! \sum_{i=0}^{n}\frac{(-1)^i}{i!}$

here $n =5.$

$\implies\ 5! \left [ \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!}+ \frac{(-1)^4}{4!} + \frac{(-1)^5}{5!} \right ]$

$= 5! \left [ 1 - 1 +\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right ]$

$= \left [ \frac{5!}{2!} - \frac{5!}{3!} + \frac{5!}{4!} - \frac{5!}{5!} \right ]$

$= 60-20+5-1$

$= 44$

Comments

Derangements

Answer 2

Here's an intutive explanation:

Suppose you want to find the no.of derangements of n then:

T(n) = total possible permutations - ( let 1 element be at it's proper place and rest are deranged + let 2 elements be at their proper place and rest are deranged + ......+ except one all elements at their proper place which is same as saying that all elements at their proper place)

T(n) = n! - (ⁿC₁ * T(n-1) + ⁿC₂ * T(n-2) + ... + 1)

Applying the eqn here

T(2) = 1

T(3) = 3! - ( ³C₁*T(2) + 1)

Similarly you can find T(4) and T(5). 

Comments

Sorry did nt understand ur solution ...

Answer 3

Total permutations possible = 5!=120

Possible number of ways in which at least one ball is in cell = 5C1*4! – 5C2*3! + 5C3*2! – 5C4*1! + 1 = 76

->120-76=44

Answer 4

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 $

=$44$