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An urn contains 3 red pens,4 green pens and 6 yellow pens.The number of ways of drawing 4 pens from the urn if atleast one red pen is to be included in the draw is (All the pens are different from each other).

$\begin{array}{1 1}(A)\;500\\(B)\;505\\(C)\;510\\(D)\;\text{None of these}\end{array} $

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1 Answer

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Required number of ways =Total number of ways choosing 4 pens -number of ways of choosing 4 non-red pens
$\Rightarrow {(3+4+6)}C_4-{4+6}C_4=13C_4-10C_4$
$\Rightarrow \large\frac{13!}{4!9!}-\frac{10!}{4!6!}$
$\Rightarrow \large\frac{13\times 12\times 11\times 10\times 9!}{4\times 3\times 2\times 1\times 9!}-\frac{10\times 9\times 8 \times 7\times 6!}{4\times 3\times 2\times 1\times 6!}$
$\Rightarrow 715-210$
$\Rightarrow 505$
Hence (B) is the correct answer.
answered Jun 20, 2014 by sreemathi.v

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