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the total number of ways of selecting two number from the set {1,2,3,4,…..3n} so that their sum divisible by 3 is equal to -A. `(2n^(2)-n)/(2)`B. `(3n^(2)-n)/(2)`C. `2n^(2)-n`D. `3n^(2)-n` |
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Answer» Correct Answer - B Given number can be rearranged as ` 1,4,7,.....,3n-2 to 3lamda-2` `2,5,8,.....3n-1 to 3 lamda-1` ` 3,6,9,....,3nto 3lamda ` that means , we must take two number from last row or one number each from first and second rows , Therefore , the total number of ways is . `""^(a)C_(2)+^(a)C_(1)xx^(a)C_(1)=(n(n-1))/(2)+n^(2)` ` =(3n^(2)-n)/(2)` |
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