Animated Solution for Mathematics - Functions: Let $A=\{a,b,c\}$ and $B=\{1,2,3,4\}$. Then the number of elements in the set $C=\{f:A\to B|2\in f(A)\text{ and }f\text{ is not one-one }\}$ is

We are given two sets: $A=\{a,b,c\}$ and $B=\{1,2,3,4\}$. We aim to find the number of functions $f:A\to B$ such that $2\in \text{Range}(f)$ and the function is not one-one.

The domain $A$ has $n(A)=3$ elements, and the codomain $B$ has $n(B)=4$ elements. Since each element in $A$ has $4$ choices in $B$, the total number of possible functions is:

The Presence of Two

We require $2\in f(A)$. It is most efficient to use the complement method: subtract the functions where $2$ is never mapped to from the total number of functions.

If $2$ is excluded from the range, each element in $A$ has only $3$ choices (the set $\{1,3,4\}$). The number of such functions is:

Thus, the number of functions where $2$ is in the range is:

Excluding One-to-One Functions

We must now remove the functions from our set of $37$ that are one-one. A function is one-one if every element in $A$ maps to a unique element in $B$.

First, we calculate the total number of one-one functions from $A$ to $B$:

Next, we identify how many of these one-one functions include $2$ in their range. We calculate the number of one-one functions that exclude $2$ by mapping $A$ to the remaining $3$ elements in $B$:

The number of one-one functions that do include $2$ in their range is:

We take our restricted universe of $37$ functions (where $2$ is in the range) and subtract the $18$ functions that are one-one.

The number of functions satisfying both conditions is:

The final answer is 19.