Numbers 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 a right-angled triangle is

none of these

Number of ways of selecting three numbers one on each card = $100 × 100 × 100 = 100^3$.

We know that $(2n+1), (2n^2 + 2n)$ and $(2n^2 + 2n+1)$ are Pythagorean triplets. Therefore, for n = 1, 2, 3, 4, 6 we get the lengths of three sides of a right angled triangle such that its hypotenuse is less than or equal to 100 cm.

Now, one Pythagorean triple (e.g. 3, 4, 5; 5, 12, 13 etc.) can be chosen in 3! ways. Therefore, the number of ways of selecting 6 Pythagorean triplets = 6 × 3!

Hence, required probability = $\frac{6 × 3!}{100^3}$