Three different numbers are selected at random from the set A = {1, 2,..., 10}. The probability that the product of two of the numbers is equal to third is

$\frac{1}{40}$

Out of 10 numbers, three numbers can be chosen in ${^{10}C}_3$ ways.

∴ Total number of elementary events = ${^{10}C}_3$

The product of two numbers, out of the three chosen numbers, will be equal to the third number if the numbers are chosen in one of the following ways:

(2, 3, 6), (2, 4, 8), (2, 5, 10)

∴ Favourable number of elementary events = 3

Hence, required probability $=\frac{3}{^{10}C_3}=\frac{1}{40}$