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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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The number of selecting atleast 4 candidates from 8 candidates is

$\begin{array}{1 1}(A)\;270\\(B)\;70\\(C)\;163\\(D)\;\text{None of these}\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Total candidates =8
Selection of the candidates =4
Required no. of selections =$8C_4+8C_5+8C_6+8C_7+8C_8$
$\Rightarrow 70+56+28+8+1$
$\Rightarrow 163$
Hence (C) is the correct answer.
answered Jun 20, 2014 by sreemathi.v

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