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64

Total white balls =2

Total black balls =3

Total red balls =4

The possibilities to draw three balls with atleast one black ball =(1 black+2 non black)+(2 black+1 non black)+(3 black+0 non black)

3C1×6C2+3C2×6C1+3C3×6C0

⇒45+18+1

⇒64

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