Question

Five different objects 1, 2, 3, 4, 5 are distributed randomly in 5 places marked 1, 2, 3, 4, 5. One arrangement is picked at random. The probability that in the selected arrangement, none of the object occupies the place corresponding to its number is:

• A. $\frac{119}{120}$
• B. $\frac{1}{5}$
• C. $\frac{11}{30}$
• D. $None$

Answer 1

The correct option is C $\frac{11}{30}$

Let objects 1, 2, 3, 4, 5 be placed in places marked 1, 2, 3, 4, 5 resp. Then the number of derangements in which none of the object occupies its original position is given by
$5![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}]$
=60-20+5-1=44
Also total number of arrangements
= 5! = 120.
Hence required probability =$\frac{44}{120}=\frac{11}{30}$