Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • $n!=n(n-1)(n-2)(n-3).....(3)(2)(1)$
  • $nP_r=\large\frac{n!}{(n-r)!}$
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 $3\times 4P_3$
$\Rightarrow 4\times 3\times 2\times 1$
$\Rightarrow 24$
$3\times 4P_3=3(24)$
$\Rightarrow 72$
Total no of ways =$120+72=192$
Hence (B) is the correct answer.
answered May 15, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App