Mathematics - Functions Question with Solution | TestHub

Total cases will be  ${}^5{C}_4\times 4!$

Now favourable cases for $2f\left(b\right)=f\left(c\right)+f\left(d\right)-f\left(a\right)$ will be,

Case (I) if $f\left(b\right)=1$ then $f\left(c\right), f\left(d\right), f\left(a\right)$ can take the value  $3,4,5$ & $4,3,5$ respectively

Case (II) if $f\left(b\right)=2$ then $f\left(c\right), f\left(d\right), f\left(a\right)$ can take value  $3,5,4 \& 5,3,4$ respectively

Case (III) if $f\left(b\right)=3$ then $f\left(c\right), f\left(d\right), f\left(a\right)$ can take value $2,5,1 \& 5,2,1$ respectively

So total favourable case will be $6$

So probability$=\frac{\mathrm{favourable} \mathrm{cases}}{\mathrm{total} \mathrm{outcomes}}$$=\frac{6}{5\times 4!}=\frac{1}{20}$