Question

Four different objects 1, 2, 3, 4 are distributed at random in four places marked 1, 2, 3, 4. What is the probability that none of the objects occupy the place corresponding to its number ?

• A. $\frac{17}{24}$
• B. $\frac{3}{8}$
• C. $\frac{1}{2}$
• D. $\frac{5}{8}$

Answer 1

Answer: (B)

$\frac{3}{8}$

Answer 2

The Number of derangement's of set with n elements is

$Dn = n![1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - .............+ (-1)^n \frac{1}{n!}]$

Probability of derangement = $\frac{Dn}{n!}$

From the case given above the value of $n$ = 4

So, The Probability that none of the objects occupy the place corresponding to their number

= $1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}$

= $1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}$

= $\frac{12 - 4 + 1}{24}$ = $\frac{9}{24}$

= $\frac{3}{8}$

The correct option is (B): $\frac{3}{8}$