Let $N$ denote the set of all positive integers. Consider the sets $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5,6,7\}$. Let $S$ be the set of all functions $f:A\to B$ such that $f\left(2\right)\ne 2$ and $f\left(4\right)\ne 4$.Consider the set $T=\{f\in S:$ there exists a function $g:B\to N$ such that $g\left(f\right(x\left)\right)=2^x$ for all $x\in A\}$.

Then the number of elements in the set $T$ is .

The correct answer is 1860.

Step 1: Understand the Condition for the Existence of Function g

We are given two sets: $A=\{1,2,3,4,5\}$ with $\left|A\right|=5$ and $B=\{1,2,3,4,5,6,7\}$ with $\left|B\right|=7$.

The set $T$ consists of functions $f:A\to B$ in $S$ for which there exists a function $g:B\to N$ such that:

$g\left(f\right(x\left)\right)=2^x$ for all $x\in A$.

Let's analyze what this condition implies for $f$:

Suppose for two elements $x_1,x_2\in A$, we have $f\left(x_1\right)=f\left(x_2\right)$. Applying $g$ to both sides gives:

$g\left(f\right(x_1\left)\right)=g\left(f\right(x_2\left)\right)\Rightarrow 2^{x_1}=2^{x_2}\Rightarrow x_1=x_2$.

This means that $f$ must be a one-one (injective) function.

Conversely, if $f$ is injective, such a function $g$ can always be defined. Therefore, the set $T$ consists of all injective functions $f:A\to B$ such that $f\left(2\right)\ne 2$ and $f\left(4\right)\ne 4$.

Step 2: Calculate the Total Number of Injective Functions

The total number of one-one (injective) functions from set $A$ (5 elements) to set $B$ (7 elements) is given by:

$N\left(\text{Total Inj}\right)={}_{7}P_5=7\times 6\times 5\times 4\times 3=2520$.

Step 3: Apply Principle of Inclusion-Exclusion (PIE)

Let:

• $E_1$ be the condition that $f\left(2\right)=2$.

• $E_2$ be the condition that $f\left(4\right)=4$.

We need to find the number of injective functions where neither $E_1$ nor $E_2$ holds:

$\left|T\right|=N\left(\text{Total Inj}\right)-N\left(E_1\right)-N\left(E_2\right)+N(E_1\cap E_2)$.

1. Calculating $N\left(E_1\right)$: Number of injective functions with $f\left(2\right)=2$.

Since $f\left(2\right)$ is fixed to $2$, the remaining 4 elements of $A$ can be mapped to any of the remaining 6 elements in $B$:

$N\left(E_1\right)={}_{6}P_4=6\times 5\times 4\times 3=360$.

2. Calculating $N\left(E_2\right)$: Number of injective functions with $f\left(4\right)=4$.

Similarly, fixing $f\left(4\right)=4$ leaves 4 elements of $A$ to map to 6 elements of $B$:

$N\left(E_2\right)={}_{6}P_4=6\times 5\times 4\times 3=360$.

3. Calculating $N(E_1\cap E_2)$: Number of injective functions with both $f\left(2\right)=2$ and $f\left(4\right)=4$.

Fixing both outputs leaves the remaining 3 elements of $A$ to map to 5 remaining elements of $B$:

$N(E_1\cap E_2)={}_{5}P_3=5\times 4\times 3=60$.

Step 4: Final Substitution

Substitute all values into the inclusion-exclusion formula:

$\left|T\right|=2520-360-360+60$.

$\left|T\right|=2520-720+60=1860$.

Thus, the total number of elements in the set $T$ is 1860.