If 3 number are chosen at random from {1,2,3,....,20}, then the probability that

$\mathrm{n}\left(\mathrm{s}\right){=}^{20}{\mathrm{C}}_3$

Number of A.P’s with c/d’s 1,2,3,….,9
Are respectively 18,16,14,…..,9. Total no.of
A.P’s = 2 + 4 + 6 + .... + 18 = 2( 1 + 2 + 3+ .... + 9) = 90

Sum = even $\Rightarrow$ (3 nos. are even) or (1 even, 2 odd)${=}^{10}{\mathrm{C}}_3+\left({ }^{10}{\mathrm{C}}_1{+}^{10}{\mathrm{C}}_2\right)=120+45+10=175$

$\mathrm{probability}(\mathrm{product} \mathrm{is} \mathrm{odd})=\left(\frac{{ }^{10}{\mathrm{C}}_3}{{ }^{20}{\mathrm{C}}_3}\right)=\frac{2}{19}$