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Two numbers $a$ and $b$ are chosen at random from the set of first 30 natural numbers.Find the probability that $a^2-b^2$ is divisible by $3$.

Total number of ways of choosing two numbers out of 1,2,3....30 is $\binom{30}{2}=435$.So total cases are 435.

But i could not count the favourable number of cases and hence the probability.Please help me.

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Note that $a^2-b^2$ is divisible by $3$ if and only if either (i) $a$ and $b$ are both divisible by $3$ or (ii) neither $a$ nor $b$ is divisible by $3$. This is because if $n$ is not divisible by $3$, then $n$ has remainder $1$ or $2$ on division by $3$. If $a$ and $b$ have the same remainder on division by $3$, then $3$ divides $a-b$. And if one has remainder $1$ and the other $2$, then $3$ divides $a+b$. Finally, $a^2-b^2=(a-b)(a+b)$.

There are $\binom{10}{2}$ choices of Type (i), and $\binom{20}{2}$ of Type (ii).

Alternately, we count the bad pairs, where one of the numbers is divisible by $3$ and the other is not. There are $(10)(20)$ bad pairs.

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  • $\begingroup$ Why are the bad pairs counted,they will not be used in finding the required probability.@Andre Nicolas $\endgroup$ – Brahmagupta Sep 29 '15 at 14:59
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    $\begingroup$ They can be used. There are $\binom{30}{2}$ pairs, so there are $\binom{30}{2}-(10)(20)$ good pairs. $\endgroup$ – André Nicolas Sep 29 '15 at 15:07

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