Question

A box contains two white balls, three black balls, and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

Answer 1

Out of $2$ white, $3$ black and $4$ red balls, three balls have to be drawn.
If at least one black ball is selected, then we have following cases:

Black. balls $\left(3\right)$	$White+redballs\left(6\right)$	Number of ways of selection
$1$	$2$	${}^3{C}_1{\times }^6{C}_2=45$
$2$	$1$	${}^3{C}_2{\times }^6{C}_1=18$
$3$	$0$	${}^3{C}_3{\times }^6{C}_0=1$
	Total	$64$