Correct Answer: $a<b<c$

Solution : Given: $a=\sqrt{6}–\sqrt{5}$, $b=\sqrt{5}–2$, $c=2–\sqrt{3}$
Rationalising the given values,
$a=\sqrt{6}–\sqrt{5}×\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$
⇒ $a=\frac{1}{\sqrt{6}+\sqrt{5}}$ -----(1)
$b=\sqrt{5}–2×\frac{\sqrt{5}+2}{\sqrt{5}+2}$
⇒ $b=\frac{1}{\sqrt{5}+2}$
⇒ $b=\frac{1}{\sqrt{5}+\sqrt{4}}$------(2)
$c=2–\sqrt{3}×\frac{2+\sqrt{3}}{2+\sqrt{3}}$
⇒ $c=\frac{1}{2+\sqrt{3}}$
⇒ $c=\frac{1}{\sqrt{4}+\sqrt{3}}$------(3)
By comparing $(a,b,c)$ we can find $a<b<c$.
Hence, the correct answer is $a<b<c$.