SSC CHSL 20 Dec 2015 Shift 2 Paper

Different combination of 2 different coloured balls
= (1 black, 1 white) , (1 black, 1 red) and (1 white, 1 red)
Number of ways of selecting $\prime \mathit{p}\prime$ Iitems out of $\prime \mathit{n}\prime$ = ${{\mathit{C}}^{\mathit{n}}}_{\mathit{p}}=\frac{\mathit{n}\left(\mathit{n}-1\right)\left(\mathit{n}-2\right)....\mathit{p}\mathrm{times}}{1\times 2\times 3\times ....\times \mathit{p}}$
⇒ Total ways toselect 2 balls of different colours = $\left({{\mathit{C}}^5}_1\times {{\mathit{C}}^3}_1\right)+\left({{\mathit{C}}^5}_1\times {{\mathit{C}}^2}_1\right)$ $+\left({{\mathit{C}}^3}_1\times {{\mathit{C}}^2}_1\right)$
= (5 × 3) + (5 × 2) + (3 × 2)
= 15 + 10 + 6 = 31