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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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From 4 officers and 8 jawans,a committee of 6 is to be chosen to include exactly one officer.The number of such committees is

$\begin{array}{1 1}(A)\;160\\(B)\;125\\(C)\;224\\(D)\;175\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Total no of officers =4
Total no of jawan =8
Total no of committees =$4C_1\times 8C_5$
$4C_1=\large\frac{4!}{1!3!}$$=4$
$8C_5=\large\frac{8!}{5!3!}=\frac{8\times 7\times 6\times 5!}{5!\times 3\times 2}$$=56$
$\Rightarrow 4\times 56$
$\Rightarrow 224$
Hence (C) is the correct answer.
answered May 19, 2014 by sreemathi.v
 

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