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# A box contain 2 white balls,3 black balls and 4 red balls.The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is

$\begin{array}{1 1}(A)\;60\\(B)\;62\\(C)\;64\\(D)\;66\end{array}$

Can you answer this question?

Toolbox:
• $c(n,r)=\large\frac{n!}{r!(n-r)!}$
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)
$\Rightarrow 3C_1\times 6C_2+3C_2\times 6C_1+3C_3\times 6C_0$
$\Rightarrow 45+18+1$
$\Rightarrow 64$
Hence (C) is the correct answer.
answered May 16, 2014