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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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A candidate is required to answer 7 questions out of 12 questions,which are divided into two groups,each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.

$\begin{array}{1 1}(A)\;780\\(B)\;782\\(C)\;784\\(D)\;786\end{array} $

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1 Answer

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
No of questions in part I=6
No of questions in part II=6
The different ways of doing the questions are
$\Rightarrow (6C_3\times 6C_4)+(6C_4\times 6C_3)+(6C_2\times 6C_5)+(6C_5\times 6C_2)$
$6C_3=\large\frac{6!}{3!\times 3!}$
$\Rightarrow \large\frac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 3\times 2\times 1}$
$\Rightarrow 20$
$6C_4=\large\frac{6!}{4!\times 2!}$
$\Rightarrow \large\frac{6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 2\times 1}$
$\Rightarrow 15$
$6C_2=\large\frac{6!}{2!\times 4!}$
$\Rightarrow \large\frac{6\times 5\times 4!}{2\times 4!}$
$\Rightarrow 15$
$6C_5=\large\frac{6!}{5!}$
$\Rightarrow \large\frac{6\times 5!}{5!}$
$\Rightarrow 6$
$\Rightarrow (20\times 15)+(15\times 20)+(15\times 6)+(6\times 15)$
$\Rightarrow 300+300+90+90$
$\Rightarrow 780$
Hence (A) is the correct answer.
answered May 15, 2014 by sreemathi.v
 

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