Question

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

Answer 1

Step 1: Interpret the data

Given,

1. $A=\left\{a,b,c\right\}$ and $B=\left\{1,2,3,4\right\}$.
2. $C$ is the set that contains all the possible output of a function $f$ whose domain is set $A$ and whose range is contained by the set $B$ such that $2$ is always in the range of $f$.
3. $f$ is not one-one which means that $f$ may produce the same output for two different inputs i.e., it is a many-one function.

Step 2: Case-1

If $f\left(x\right)=2\forall x\in A$ i.e., for all $x$ belonging to $A$.

There can only be one function such that it maps all elements of a set to a single element of another set.

Here, $f$ is a constant function since it maps all its input into a single output.

Thus, from this case, we have $1$ function.

Step 3: Case-2

If $f\left(x\right)=2$ for exactly two of the elements.

Then we have that, of the three elements, two map to $2$. This is in ${}^{3}C_2$ ways.
Then the remaining one maps to one of the other three. This is in ${}^{3}C_1$ (The remaining three in $B$ choose one from $A$).

Thus, the possible number of functions is
$\begin{array}{rcl}{}^{3}C_2\times {}^{3}C_1& =& \frac{3!}{\left(3-2\right)!\times 2!}\times \frac{3!}{\left(3-1\right)!\times 1!}\\ & =& \frac{6\times 6}{2\times 2\times 1}\\ & =& 9\end{array}$

Step 4: Case-3

If $f\left(x\right)=2$ for exactly one of the elements.

Then we have that, of the three elements, one maps to $2$. This is in ${}^{3}C_1$ ways.
Then the remaining two map to one of the other three. This is in ${}^{3}C_2$ (The remaining two in $B$ choose two from $A$).

Thus, the possible number of functions is
$\begin{array}{rcl}{}^{3}C_1\times {}^{3}C_2& =& \frac{3!}{\left(3-1\right)!\times 1!}\times \frac{3!}{\left(3-2\right)!\times 2!}\\ & =& \frac{6\times 6}{2\times 2\times 1}\\ & =& 9\end{array}$

So, total number of possible functions is $1+9+9=19$.

Hence the total number of possible functions from $A$ to $B$ with given conditions is $19$.