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JEE MATHEMATICS MCQ: PERMUTATIONS

1 Hard

1 of 1

Application +4 | -1

In an examination, the maximum marks for each of three papers is

n

and that for fourth paper is

2 n

. Then, the number of ways in which a candidate can get

3 n

marks, is

View answer and explanation

Correct option: B

\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right)

Option B (

\frac{{1}}{{6}}(n+1)\left(5 n^2+10 n+6\right)

) is correct because it represents the number of ways a candidate can get 3n marks, considering the maximum marks for each paper.

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