Number 1, 2, 3, ..., 100 are written down on each of the cards A, B and C. One number is selected at random from each of the cards. The probability that the numbers so selected can be the measures (in cm) of three sides of right angled triangles no two of which are similar, is

$\mathrm{n}\left(\mathrm{S}\right)=100\times 100\times 100$

We know that $(2\mathrm{n}+1{)}^2+{\left(2{\mathrm{n}}^2+2\mathrm{n}\right)}^2={\left(2{\mathrm{n}}^2+2\mathrm{n}+1\right)}^2$ for all $\mathrm{n}\in \mathrm{N}$

sides of a right angled triangle whole longest side $\le$100.
e.g. When n = 1, sides are 3, 4, 5
and when n = 2, sides are 5, 12, 13 and so on'
The number of selections of 3, 4, 5 from the three cards by taking one from each is 3!. 

$\therefore \mathrm{n}\left(\mathrm{E}\right)=6(3!)$

Hence, $\mathrm{P}\left(\mathrm{E}\right)=\frac{6(3!)}{100\times 100\times 100}=\frac{1}{100}{\left(\frac{3}{50}\right)}^2$