Tough one !
In an examination held by Sherwood Boarding School, the maximum marks for each of three papers is n and that for the fourth paper is 2n. Find the number of ways in which a candidate can get 3n marks.
Ans. (n+1)(5n^{2}+10n+6)
I'm stumped on this one. I tried hard but was unable to solve this algebraically . How should we approach this algebraically ? Request an algebraic solution.
Thanks in advance !
Hi Idea,
I have tried to solve your expression but I am not getting an answer (n+1)(5n^{2}+10n +6) . Maybe there is something missing in the equation. Please advise how to get to the solution and what should be the correct equation.
'my' expression is the answer to 'my' problem where a,b,c,d >= 0
'your' expression is the answer to 'your' problem where 0 <= a,b,c <= n and 0 <= d <= 2n
You can subtract some things from my answer to get to your answer
What things? solutions for which a > n or b > n or c > n or d > 2n
That's really way over my head. I really can't relate that to a + b + c + d = 3n with the given constraints for a,b,c and d. Maybe I'm not ready for this yet. Can't help, haven't done any math beyond my 12th grade. Don't have any college math background.