How many ordered pairs?

$a*b = a + b$

$\Rightarrow a*b - a -b + 1 = 1$

$\Rightarrow (a-1)(b-1) = 1$

The only solutions are $(a-1,b-1) = (1,1),(-1,-1) \Rightarrow (a,b) = (2,2),(0,0)$ . But in both of them, $a = b$ . Therefore there are no solutions.

Because we are given that $a\times b=a+b$ we can see that $a\times{(b-1)}=b.$ From this,we get that $a=\dfrac{b}{b-1}.$ Now, $b$ and $b-1$ are co-prime integers.Thus, $\dfrac{b}{b-1}$ would never be an integer.But,if we can get the denominator equal to $1$ then we are done,so $b-1=1\Rightarrow b=2.$ But if $b=2$ then $a=2$ but this can't happen as it is given in the question that $a\neq b.$ Thus, $0$ solutions.

$a + b = ab \Rightarrow a - ab + b = 0 \Rightarrow a\times{(1-b)} = -b \Rightarrow a = \dfrac{b}{(b-1)}$ .

Since $a$ has to be an integer.

$\therefore \dfrac{b}{(b-1)}$ must be an integer. Hence b should be divisible by $b - 1$ and the only possible value of $b$ is $2$ which gives $\boxed{a = 2}$ .

But according to the question $a! = b \Rightarrow a = b = 2$ .

So there is no possible pair.