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. Then find 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.

Options

A. `(4)/(100^(3))`
B. `(3)/(50^(3))`
C. `(3!)/(100^(3))`
D. none of these

Answer

D

Explanation

Number of ways of selecting three numbers one on each card =`100xx100xx100=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 `=6xx3!`
Hence, required probability `=(6!xx3!)/(100^(3))`