$\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

$151=1+(m-1)\times 2$

$m-1=\large\frac{151-1}{2}$

$m=76$

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.

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