AP EAPCET 05-Jul-2022 Shift 1 Paper

Total marks = Marks for first 3 papers + Marks for fourth paper =3n+2n=5n
Candidate needs to get 3n marks.
Let x1,x2,x3,x4 be the marks of candidate in I, II, III and IV paper, respectively.
Then, x1+˙x2+x3+x4
=3n(0≤x1,x2,x3,≤n‌ and ‌0≤x4≤2n)
Using multinomial theorem,
Number of ways = Coefficient of x3n in
‌

(1+x+...+xn)3

╰───────▾───────╯ ‌GP series ‌

(1+x+...x2n)

╰──────▾──────╯ ‌GP series ‌

‌⇒(‌

1−xn+1

1−x

)3(‌

1−x2n+1

1−x

)
Finding the coefficient of x3n using binomial expansion, we get
‌3n+3C3−32n+2C3+3n+2C3−‌n+3C3
=‌

1

6

(n+1)(5n2+10n+6)