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

Step-by-step solution

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 x₁, x₂, x₃, x₄ be the marks of candidate in I, II, III and IV paper, respectively. Then, x₁+x₂+x₃+x₄=3 n (0 x₁, x₂, x₃, n . and .0 x₄ 2 n ) Using multinomial theorem, Number of ways = Coefficient of x^ 3 n in (1+x+ .+x^n )^3 _ GP series (1+x+ x^ 2 n ) _ GP Series aligned & 1-x^ 3(n+1) -3 x^ n+1 (1-x^ n+1 ) (1-x^ 2 n+1 )(1-x)⁻⁴ & (1-x^ 3(n+1) -3 x^ n+1 +3 x^ 2 n+2 ) (1-x^ 2 n+1 )(1-x)⁻⁴ aligned aligned