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

The Correct Option is B

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): \(\frac{3}{8}\)