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 $f\left(x_i\right)=x_i$ for exactly four of the $x_i$ is

Answer 1

Given : $f\left(x_i\right)=x_i$ for exactly four elements out of six.

$\Rightarrow$ Way of selecting such four elements $={}^6{C}_4$
And the rest two elements have two way of mapping i.e., $(x_5\to x_5\&x_6\to x_6$ or $x_5\to x_6\&x_6\to x_5)$


But, it is given only four elements satisfy $f\left(x_i\right)=x_i,$ so one way get cancelled out.
$\therefore$ Total number of ways $=1\cdot {}^6{C}_4=15$