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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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In an examination a question paper consists of 12 questions divided into two parts (i.e) Part I and Part II.Containing 5 and 7 questions respectively.A student is required to attempt 8 question in all,selecting at least 3 from each part.In how many ways can a student select the questions 2

$\begin{array}{1 1}(A)\;320\\(B)\;420\\(C)\;520\\(D)\;620\end{array} $

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
  • $n!=n(n-1)(n-2)(n-3).......(3)(2)(1)$
Students may select 8 questions according to the following scheme.
If P is the required number of ways,then
$P=C(5,3)\times C(7,5) +C(5,4)\times C(7,4)+C(5,5)\times C(7,3)$
$\;\;\;=\large\frac{5!}{3!2!}\times \frac{7!}{5!3!}+\frac{5!}{4!1!}\times \frac{7!}{4!3!}+\frac{5!}{5!0!}\times \frac{7!}{3!4!}$
$\;\;\;=\large\frac{5\times 4 \times 3!}{3!\times 2\times 1}\times \frac{7\times 6\times 5!}{5!\times 3\times 2\times 1}+\frac{5\times 4!}{4!}\times \frac{7\times 6\times 5\times 4!}{4!\times 3\times 2\times 1}+ \frac{5\times 7\times 6\times 5\times 4!}{3\times 2\times 1\times4!}$
$\;\;\;=10\times 7+5\times 35+5\times 35$
$\;\;\;=70+175+175$
$\;\;\;=420$ ways
Hence (B) is the correct answer.
answered May 14, 2014 by sreemathi.v
 

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