RRB NTPC Mains or Stage 2 Model Paper 4 with solutions

(c) There are 5 + 7 = 12 balls in the bag and out of these two balls can be drawn in ${}_{}^{12}{\mathit{C}}_2^{}$ ways. There are 5 green balls, therefore, one green ball can be drawn in $5\mathit{C}1$ ways; similarly, one red ball can be drawn in $7\mathit{C}1$ ways so that the number of ways inwhich we can draw one green ball and the other red is ${}_{}^5{\mathit{C}}_1^{}{\times }^7{\mathit{C}}_{1.}$Hence, P (one green and the other red)
$\frac{{}_{}^5{\mathit{C}}_1^{}{\times }^7{\mathit{C}}_1}{{12}_2^{\mathit{C}}}=\frac{5}{1}\times \frac{7}{1}\times \frac{1\times 2}{12\times 11}=\frac{35}{66}$