Let S = {a, b, c} and T = {1, 2, 3}.

S = {a, b, c}, T = {1, 2, 3}

(i) F: S → T is defined as:

F = {(a, 3), (b, 2), (c, 1)}

$\Rightarrow F(a)=3, F(b)=2, F(c)=1$

Therefore, $\mathrm{F}^{-1}: T \rightarrow S$ is given by

$\mathrm{F}^{-1}=\{(3, a),(2, b),(1, c)\}$

(ii) $F: S \rightarrow T$ is defined as:

$F=\{(a, 2),(b, 1),(c, 1)\}$

Since F (b) = F (c) = 1, F is not one-one.

Hence, $\mathrm{F}$ is not invertible i.e., $\mathrm{F}^{-1}$ does not exist.