AP EAMCET · Maths · Permutation Combination
In an examination, the maximum marks for each of three subjects is \(n\) and that for the fourth subject is \(2 n\). The number of ways in which candidates can get \(3 n\) marks is
- A \(\frac{1}{6}(n+1)^2\left(5 n^2+10 n+6\right)^2\)
- B \(\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right)^2\)
- C \(\frac{1}{6}(n+1)^2\left(5 n^2+10 n+6\right)\)
- D \(\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right)\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right)\)
Step-by-step Solution
Detailed explanation
Total marks \(=\) Marks for first 3 papers + Marks for fourth paper \(=3 n+2 n=5 n\) Candidate needs to get \(3 n\) marks. Let \(x_1, x_2, x_3, x_4\) be the marks of candidate in I, II, III and IV paper, respectively. Then, \(x_1+x_2+x_3+x_4\)…
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