All possible two-factor products are formed f | vedclass

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Question

(B) The total number of ways to choose two distinct factors from $200$ numbers is given by $^{200}C_2$.$^{200}C_2 = \frac{200 	imes 199}{2} = 19900$.

Vedclass · Accepted Answer

$7180$

Vedclass · Answer

(B) The total number of ways to choose two distinct factors from $200$ numbers is given by $^{200}C_2$.$^{200}C_2 = \frac{200 	imes 199}{2} = 19900$.$A$ product is $NOT$ a multiple of $5$ if both chosen factors are not multiples of $5$.The numbers from $1$ to $200$ that are multiples of $5$ are $5, 10, \dots, 200$,which is $\frac{200}{5} = 40$ numbers.The numbers that are $NOT$ multiples of $5$ are $200 - 40 = 160$.The number of products that are $NOT$ multiples of $5$ is $^{160}C_2 = \frac{160 	imes 159}{2} = 80 	imes 159 = 12720$.The number of products that $ARE$ multiples of $5$ is the total products minus those that are not multiples of $5$:$19900 - 12720 = 7180$.

All possible two-factor products are formed from the numbers $1, 2, 3, 4, \dots, 200$. The number of products out of the total obtained which are multiples of $5$ is:

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