In a box there are 5 different black balls, 3 different white balls and 2 different red balls. How many ways to select 2 Balls but not both same colour ?

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 $$'p'$$ items out of $$'n'$$ = $$C^n_p=\frac{n(n-1)(n-2)....p\ times}{1\times2\times3\times....\times p}$$

=> Total ways to select 2 balls of different colours = $$(C^5_1\times C^3_1)$$ $$+$$ $$(C^5_1\times C^2_1)$$ $$+$$ $$(C^3_1\times C^2_1)$$

= $$(5\times3)+(5\times2)+(3\times2)$$

= $$15+10+6=31$$

=> Ans - (B)