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 derangement is a permutation of objects that leave no object in its original position.Number of derangement's of set with n elements is

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

The probability of derangement = $$\frac{D_n}{n!}$$

Now, in the given case $$n=4$$

$$\therefore$$ Probability that none of the objects occupy the place corresponding to their number

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

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

= $$\frac{12-4+1}{24}=\frac{9}{24}=\frac{3}{8}$$

=> Ans - (B)