A box contains two distinct white balls, three distinct black balls and four distinct 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 drawn

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Solution

Total number of ways are $=$ (one black and any two from two white and four red balls) or (two black and any one from two white and four red balls) or (three black and zero from two white and four red balls)
$=^{3} C_{1} \times^{6} C_{2} +^{3} C_{2} \times^{6} C_{1} +^{3} C_{3} \times^{6} C_{0}$
$= 3 \times 15 + 3 \times 6 + 1$
$= 45 + 18 + 1$
$= 64$

Alternate Solution:
Total possible ways to drawn three balls $=^{9} C_{3} = 84$
When no black ball is drawn then number of ways $=^{6} C_{3} = 20$
Hence number of ways of drawing atleast one black ball is
$=$ Total possible ways $-$ No black ball drawn
$= 84 - 20 = 64$

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