1. ## Algebraic Solution

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. $\displaystyle \frac{1}{6}$(n+1)(5n2+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.

Tough one !

3. ## Re: Algebraic Solution

How many solutions does the following system have

$\displaystyle a+b+c+d = 3n$

$\displaystyle 0\leq a \leq n$

$\displaystyle 0\leq b \leq n$

$\displaystyle 0\leq c \leq n$

$\displaystyle 0\leq d \leq 2n$

where a, b, c, and d are integers? (Discrete Math / Combinatorics)

4. ## Re: Algebraic Solution

Thanks Idea! That was cool. So it is 8n-1C5n-1. Am I right ?

This comes out to be $\displaystyle \frac{(8n-1)!}{3n!(5n-1)!}$. But how do we solve from there ? Please advise .

5. ## Re: Algebraic Solution

No, that's not right.

How many solutions does the following system have

$\displaystyle a+b+c+d=3n$

where $\displaystyle a,b,c,d \geq 0$?

Answer: $\displaystyle C(3n+3,3)=\frac{(3n+3)!}{(3n)!3!}$

6. ## Re: Algebraic Solution

Hi Idea,

I have tried to solve your expression $\displaystyle \frac{(3n+3)!}{(3n)!3!}$ but I am not getting an answer $\displaystyle \frac{1}{6}$(n+1)(5n2+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.

7. ## Re: Algebraic Solution

'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

What things? solutions for which a > n or b > n or c > n or d > 2n

8. ## Re: Algebraic Solution

Hi Idea,

I have tried 3n+2C3 and 2n+3C3 but unable to get to the correct answer. Please elaborate how to form the correct equation and solve to reach the correct answer .

9. ## Re: Algebraic Solution

Try this

$\displaystyle C(3n+3,3)-3C(2n+2,3)-C(n+2,3)+3C(n+1,3)=$

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

10. ## Re: Algebraic Solution

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.