From 3n consecutive integers, three integers | vedclass

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Question

(C) Let the $3n$ consecutive integers be partitioned into three sets based on their remainder when divided by $3$: $G_1 = \{x, x+3, \dots, x+3(n-1)\}$

Vedclass · Accepted Answer

$\frac{3n^2-3n+2}{(3n-1)(3n-2)}$

Vedclass · Answer

(C) Let the $3n$ consecutive integers be partitioned into three sets based on their remainder when divided by $3$: $G_1 = \{x, x+3, \dots, x+3(n-1)\}$ $G_2 = \{x+1, x+4, \dots, x+3(n-1)+1\}$ $G_3 = \{x+2, x+5, \dots, x+3(n-1)+2\}$ Each set contains $n$ integers. For the sum of three integers to be divisible by $3$, either all three must be from the same set, or one must be chosen from each of the three sets. Number of ways to choose $3$ from the same set: $3 	imes \binom{n}{3} = 3 	imes \frac{n(n-1)(n-2)}{6} = \frac{n(n-1)(n-2)}{2}$. Number of ways to choose one from each set: $\binom{n}{1} 	imes \binom{n}{1} 	imes \binom{n}{1} = n^3$. Total favorable ways: $\frac{n(n-1)(n-2)}{2} + n^3 = \frac{n(n^2-3n+2) + 2n^3}{2} = \frac{3n^3-3n^2+2n}{2}$. Total sample space: $\binom{3n}{3} = \frac{3n(3n-1)(3n-2)}{6} = \frac{n(3n-1)(3n-2)}{2}$. Probability: $\frac{3n^3-3n^2+2n}{n(3n-1)(3n-2)} = \frac{3n^2-3n+2}{(3n-1)(3n-2)}$.

From $3n$ consecutive integers, three integers are selected at random. The probability that their sum is divisible by $3$ is

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