Question

A is a set of four elements. A bijection on A is selected at random. The probability that it satisfies the condition
$f\left(x\right)\ne x$ for all $x\in A$ is

• A. $\frac{17}{24}$
• B. $\frac{15}{24}$
• C. $\frac{11}{24}$
• D. $\frac{9}{24}$

Answer 1

Answer: (D)

$\frac{9}{24}$

Suppose $A$ be the set with elements $\{x_1,x_2,x_3,x_4\}$.
The number of bijections that can be formed on set $A$ is equal to $4\times 3\times 2\times 1=24$.
Hence, $24=$ (cases where $f\left(x\right)=x$ for at least one $x$) $+$ (cases where $f\left(x\right)\ne x$ for any $x$)
Lets find all cases where $f\left(x\right)=x$ for at least one value of $x$.
It includes 3 cases:
1. $f\left(x\right)=x$ for exactly one value of $x$.
Suppose $f\left(x_1\right)=x_1$. Then $f\left(x_2\right)=x_{3}or{x}_4$. So two cases possible. Similarly, two cases are possible for each of $f\left(x_2\right)=x_2,f\left(x_3\right)=x_3,f\left(x_4\right)=x_4$. Hence total number of cases $=4\times 2=8$
2. $f\left(x\right)=x$ for exactly two values of $x$.
Suppose $f\left(x_1\right)=x_1$ and $f\left(x_2\right)=x_2$. Then only one case possible: $f\left(x_3\right)=x_4$ and $f\left(x_4\right)=x_3$
Any two elements of $A$ can be chosen in ${}^4{C}_2$ ways.
So, the total number of cases ${=}^4{C}_2\times 1=6$.
3. $f\left(x\right)=x$ for all values of $x$. Only one case possible here.
Hence total number of cases where $f\left(x\right)=x$ for at least one value of $x$ is equal to $8+6+1=15$.
So , (cases where $f\left(x\right)\ne x$ for any $x$) $=24-15=9$
Probability$=\frac{9}{24}$