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out of 3n consecutive number . find number of way in which 3 number can be selected such that their sum is divisble by 3
n numbers will leave remainder 0 when divided by 3, n will leave remainder 1, and n will leave remainder 2,
a set of 3 numbers will have a sum divisible by 3 if: let (x,y,z) be the set of remainders when divided by 3,
we require (x,y,z)=(0,0,0),(0,1,2),(1,1,1), (2,2,2), so total number of ways= (n(n-1)(n-2))*4=4n(n-1)(n-2)
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