a = $$\sqrt 6 \,$$ - $$\sqrt 5 $$ , b = $$\sqrt 5 \,$$ - 2, c = 2 - $$\sqrt 3 $$  then point out the correct alternative among the four alternatives given below ?

$$\eqalign{ & a = \sqrt 6 - \sqrt 5 \cr & b = \sqrt 5 - 2 \cr & c = 2 - \sqrt 3 \cr & {\text{Rationalize all the terms}} \cr & a = \frac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} \times \sqrt 6 + \sqrt 5 \cr & \,\,\,\,\, = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}} \times \sqrt 5 + 2 \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + 2}} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} \times 2 + \sqrt 3 \cr & \,\,\,\,\,\, = \frac{1}{{2 + \sqrt 3 }} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr & {\text{Now, }} \cr & a = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr} $$
Now the term whose denominator is largest is the smallest term.
So, a < b < c