Three different numbers are selected at random from the set $$A = \left\{ {1,\,2,\,3,.....,10} \right\}.$$      The probability that the product of two of the numbers is equal to the third is :

Solution :
$$n\left( S \right) = {}^{10}{C_3}.$$   Clearly, $$E = \left\{ {\left( {10,\,5,\,2} \right),\,\left( {8,\,2,\,4} \right),\,\left( {6,\,2,\,3} \right)} \right\}.\,{\text{So, }}n\left( E \right) = 3$$
$$\therefore \,P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{3}{{{}^{10}{C_3}}} = \frac{1}{{40}}.$$