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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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Find the number of integers greater than 7000 that can be formed with the digits 3,5,7,8 and 9 where no digits are repeated

$\begin{array}{1 1}(A)\;191\\(B)\;192\\(C)\;193\\(D)\;194\end{array} $

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1 Answer

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  • $n!=n(n-1)(n-2)(n-3).....(3)(2)(1)$
  • $nP_r=\large\frac{n!}{(n-r)!}$
Digits are 3,5,7,8,9
Besides 4 digits integers greater than 7000 five digits integers are always greater than 7000
The number of such integers is $5P_5=5!$
$5!=5\times 4\times 3\times 2\times 1$
$\Rightarrow 120$ ways
For a four digit integer to be greater than 7000 it must begin with 7,8 or 9.The number of such integer is $3\times 4P_3$
$4P_3=\large\frac{4!}{(4-3)!}=\frac{4!}{1!}$
$\Rightarrow 4\times 3\times 2\times 1$
$\Rightarrow 24$
$3\times 4P_3=3(24)$
$\Rightarrow 72$
Total no of ways =$120+72=192$
Hence (B) is the correct answer.
answered May 15, 2014 by sreemathi.v
 

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