If $a\sqrt{3}+b\sqrt{3}=\sqrt{75}$ , where $a,b\in N$ , then the number of possible ordered pairs of the form $\left(a,b\right)$  can be obtained is

$a\sqrt{3}+b\sqrt{3}=5\sqrt{3}$

$\left(a+b\right)\sqrt{3}=5\sqrt{3}$

$a+b=5$

Possible ordered pairs $\left\{\left(1,4\right),\left(2,3\right),\left(3,2\right),\left(4,1\right)\right\}$  

No., of possible ordered pairs is 4.