Let R = {a,b,c,d,e}and S = {1,2,3,4}Total number of onto functions f :R → S such that f (a) $\ne$ 1, is equal to

$\text{ If }f\left(a\right)=1\mathrm{\&}\text{ one of }f\left(b\right),f\left(c\right),f\left(d\right),f\left(e\right)=1\text{ then total such cases }=4\cdot 3!=24$

$$\begin{gathered} \text{ If only }f\left(a\right)=1\text{, then Total cases } \\ =3^4-\left({ }^3{C}_1\cdot 2^4\right)+\left({ }^3{C}_2\cdot 1\right)=36 \end{gathered}$$
$$\begin{gathered} No of onto functions when f\left(a\right) = 1 is 24 + 36 = 60 \\ Total no. of onto functions \\ \begin{array}{l}=4^5-\left({ }^4{C}_1\cdot 3^5\right)+\left({ }^4{C}_2\cdot 2^5\right)-\left({ }^4{C}_3\cdot 1\right)\\ =1024-973+192-4\\ =240\end{array} \end{gathered}$$
Number of required functions = 240 – 60 = 180