The probability that a randomly chosen one-one function from the...

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 $3, 5, 4 \text{ and } 5, 3, 4$ 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 $2, 5, 1 \text{ and } 5, 2, 1$ respectively

So total favourable case will be $6$

So probability$= \dfrac{favourablecases}{totaloutcomes}$$= \dfrac{6}{5 \times 4 !} = \dfrac{1}{20}$