Question

Suppose that $4$ cards labeled $1$ to $4$ are placed randomly into $4$ boxes also labeled $1$ to $4$, one card per box. What is the probability that no card gets placed into a box having the same label as the card?

• A. $\frac{1}{3}$
• B. $\frac{3}{8}$
• C. $\frac{5}{12}$
• D. $\frac{9}{16}$

Answer 1

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

The above is a problem of dearrangement

No. of dearrangements $=n!(1-\frac{1}{1!}+\frac{1}{2!}-.........\frac{1}{n!})$

Here $n=4$

$=4!(1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24})$

$=12-4+1$

$=9$

$\Rightarrow$ Total possibilities $4!=24$

$\Rightarrow$ Probability $=\frac{9}{24}$

$\Rightarrow P=\frac{3}{8}$

Hence, the answer is $\frac{3}{8}.$