A box contain 2 white balls, 3 black balls and 4 red balls.

At-least 1 black ball can be selected in following ways 

1 black ball and two non - black or 2 black ball and one non - black or all 3 black balls

∴ Total number of ways of selecting is

$=\frac{{ }^{3} C_{3}}{\text { all black }}+\frac{{ }^{3} C_{2} \times{ }^{6} C_{1}}{2 \text { black balls }}+\frac{{ }^{3} C_{1} \times{ }^{6} C_{2}}{2 \text { black balls }}$

$=1+3 \times 6+3 \times \frac{6 \times 5}{5}$

$=1+18+45$

$=64$

i.e The number of ways three balls be drawn from the box with at-lest one black ball included is 64.