Ask Questions, Get Answers

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

Two numbers are chosen from 1,3,5,7.....147,149 and 151 and multiplied together in all possible ways.The number of ways which will give us the product a multiple of 5 is

$\begin{array}{1 1}(A)\;1710\\(B)\;2900\\(C)\;1700\\(D)\;\text{None of these}\end{array} $

Can you answer this question?

1 Answer

0 votes
  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Out of the given numbers,the numbers which are multiple of 5 are 5,10,15,20.....145
This is an AP whose first term a=5
Common difference =5
$n^{th}$ term,$t_n=145$
$145=5+(n-1)\times 5$
$\Rightarrow n=29$
Similarly if total number of numbers is $m$ then
$151=1+(m-1)\times 2$
So number of ways in which product is a multiple of 5
$\Rightarrow$ Both two numbers from 5,15,25....145+one number from 5,15,25,....145 and one from remaining numbers
$\Rightarrow 29C_2+29C_1\times 76C_1$
$\Rightarrow 29\times 14+29\times 76$
$\Rightarrow 29\times 90$
Hence (D) is the correct answer.
answered Jun 20, 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