Question

$A,B,C$ and $D$ are the sets with $3,3,4$ and $4$ elements respectively. If
$f\left(x\right)$ is a function defined from $A$ to $C$
$g\left(x\right)$ is a function defined from $D$ to $A$
$h\left(x\right)$ is a function defined from $C$ to $D$
$i\left(x\right)$ is a function defined from $A$ to $B$

Then number of functions possible for

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& h\left(x\right)\text{if it is one - one}& \left(P\right)& 36\\ \text{(2)}& g\left(x\right)\text{if it is onto}& \left(Q\right)& 40\\ \text{(3)}& i\left(x\right)\text{if it is into}& \left(R\right)& 64\\ \text{(4)}& f\left(x\right)\text{if it is many - one}& \left(S\right)& 48\\ & & \left(T\right)& 3\\ & & & \\ & & \left(U\right)& 24\end{array}$

Which of the following is the correct combination?

• A. $\left(3\right)\to \left(U\right)$
• B. $\left(3\right)\to \left(S\right)$
• C. $\left(4\right)\to \left(R\right)$
• D. $\left(4\right)\to \left(Q\right)$

Answer 1

The correct option is D $\left(4\right)\to \left(Q\right)$

$h\left(x\right)$ is one-one.
So, each element of domain will have one image in co-domain. So, first element has $4$ options, second has $3$ options and so on.
So number of one - one functions for $h\left(x\right)$ is $4!=24$

Number of functions if $g\left(x\right)$ is onto
$=\sum _{k=0}^3(-1{)}^k{}^n{C}_k(3-k{)}^4$
$=3^4-{}^3{C}_1\times (3-1{)}^4+{}^3{C}_2\times (3-2{)}^4$
$=81-48+3=36$

Number of into functions for $i\left(x\right)$
$=$ Total functions $-$ Total onto functions
Total functions, $i\left(x\right)$ is $3^3$
Total onto functions $=3!$
So, number of into functions $=9-6=3$

Number of many-one functions for $f\left(x\right)$
$=$ total functions $-$ total one-one functions
Total functions = $4^3$
Total number of one-one functions $=4\times 3\times 2=24$
So, number of many-one functions $=64-24=40$