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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.

$\begin{array}{1 1}(A)\;1550\\(B)\;1440\\(C)\;1444\\(D)\;1500\end{array} $

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1 Answer

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  • $nP_r=\large\frac{n!}{(n-r)!}$
First women can take any of the chairs marked 1 to 4 in 4 different way.
Second women can take any of the remaining 3 chairs from those marked 1 to 4 in 3 different ways.
So,total no of ways in which women can take seat =$4\times 3$
$\Rightarrow 4P_2$
$4P_2=\large\frac{4!}{(4-2)!}$
$\qquad=\large\frac{4\times 3\times 2\times 1}{2\times 1}$
$\qquad =12$
After two women are seated 6 chairs remains
First man take seat in any of the 6 chairs in 6 different ways,second man can take seat in any of the remaining 5 chairs in 5 different ways
Third man can take seat in any of the remaining 4 chairs in 4 different ways.
So,total no of ways in which men can take seat =$6\times 5\times 4$
$\Rightarrow 6P_3$
$6P_3=\large\frac{6!}{(6-3)!}$
$\Rightarrow \large\frac{6\times 5 \times 4\times 3\times 2\times 1}{3\times 2\times 1}$
$\Rightarrow 120$
Hence total number of ways in which men and women can be seated =$120\times 12$
$\Rightarrow 1440$
Hence (B) is the correct answer.
answered May 14, 2014 by sreemathi.v
 

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