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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} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
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

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