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A box contain two white,three black and four red balls. In how many ways can three balls be drawn from the box,if atleast one black ball is to be includedin the draw

$\begin{array}{1 1}(A)\;74\\(B)\;84\\(C)\;64\\(D)\;20\end{array} $

1 Answer

  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
The possibilities of choosing at least one block ball are :-
1 black +2 non-black (or) 2black +1 non-black (or) 3 black +0 non-black
$\therefore$No of ways of choosing 3 balls with at least one black ball =$(3C_1\times 6C_2)+(3C_2\times 6C_1)+(3C_3\times 6C_0)$
$3C_1=\large\frac{3!}{1!2!}=\frac{3\times 2\times 1}{2\times 1}$
$\Rightarrow 3$
$6C_2=\large\frac{6!}{2!4!}=\frac{6\times 5\times 4!}{2\times 4!}$
$\Rightarrow 15$
$3C_2=\large\frac{3!}{2!1!}=\frac{3\times 2!}{2\times 1}$
$\Rightarrow 3$
$6C_1=\large\frac{6!}{1!\times 5!}$
$\Rightarrow 6$
$3C_3=\large\frac{3!}{3!\times 0!}$
$\Rightarrow 1$
$\Rightarrow (3\times 15)+(3\times 6)+1$
$\Rightarrow 45+18+1$
$\Rightarrow 64$
Hence (C) is the correct answer.
answered May 15, 2014 by sreemathi.v

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