Let A={a, b, c} and B={1,2,3,4}.


Let $A=\{a, b, c\}$ and $B=\{1,2,3,4\}$. Then the number of elements in the set $C=\{f: A \rightarrow B \mid 2 \in f(A)$ and $f$ is not one-one $\}$ is


The desired functions will contain either one element or two elements in its codomain of which ' 2 ' always belongs to $f(A)$.

$\therefore$ The set $B$ can be $\{2\},\{1,2\},\{2,3\},\{2,4\}$

Total number of functions $=1+\left(2^{3}-2\right) 3=19$.

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