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  4. Mr. A lists all the positive divisors of the number N =(2010)2. If Mr.A randomly selects two divisors from the list then the probability that exactly one of the selected divisor is a perfect square, is

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Q. Mr. A lists all the positive divisors of the number $N =(2010)^2$.
If Mr.A randomly selects two divisors from the list then the probability that exactly one of the selected divisor is a perfect square, is

Permutations and Combinations

Solution:

$ N =(2010)^2=2^2 \cdot 3^2 \cdot 5^2 \cdot(67)^2=( a \cdot b \cdot c \cdot d \text { say }) $
$\text { Total number of divisors }=3 \cdot 3 \cdot 3 \cdot 3=81 $
$\text { Number of divisors which are perfect square is }={ }^4 C _0+{ }^4 C _1+{ }^4 C _2+{ }^4 C _3+{ }^4 C _4=16$
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Let $E$ : two randomly chosen divisors have exactly one perfect square.
$P ( E )=\frac{{ }^{16} C _1 \cdot{ }^{65} C _1}{{ }^{81} C _2}=\frac{26}{81}$