Question

Let $A=\{x_1,x_2,x_3,x_4,x_5,x_6\}$ and $f:A\to A$. The number of bijective functions such that $f\left(x_i\right)\ne x_i$ for exactly $3$ elements $(i=1\text{to}6)$ is

Answer 1

$3$ elements can be selected from $6$ elements of $A$ in ${}^6{C}_3$ ways.
For these $3$ elements, $f\left(x_i\right)\ne x_i$
So, these elements can be dearranged in $D_3=3![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}]$
Now, for remaining $3$ elements they can be arranged in $1$ way $(\because f(x_i)=x_i)$
$\therefore \text{Total required bijective functions}={}^6{C}_3\left(D3\right)$
$={}^6{C}_{3}3![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}]$
$=\frac{6\cdot 5\cdot 4}{1\cdot 2\cdot 3}\cdot 2=40$