Explanation

(c)
In 3$n$ conseecutive natural numbers, either
(i) $n$ numbers are of from $3P$
(ii) $n$ numbers are of from $3P + 1$
$(iii)n$ numbers are of from $3P + 2$
Here favourable number of cases= Either we can select three numbers from any of the set or we can select one from each set
$= \ ^{n}C_{3}{+ \ }^{n}C_{3} +^{n}C_{3} + (\ ^{n}C_{1} \times \ ^{n}C_{1}{\times \ }^{n}C_{1})$
$= 3\left( \frac{n(n - 1)(n - 2)}{6} \right) + n^{3}$
$= \frac{n(n - 1(n - 2)}{2} + n^{3}$
Total number of selections $= \ ^{3n}C_{3}$
$\therefore$Required probability
$= \frac{\frac{n(n - 1)(n - 2)}{2} + n^{3}}{\frac{3n(3n - 1)(3n - 2)}{6}}$
$= \frac{3n^{2} - 3n + 2}{(3n - 1)(3n - 2)}$