No. of ways a candidate can qualify for an exam In an entrance examination, there are two papers of 50 marks and one paper of 100 marks. A candidate will qualify to take admission if he secures at least 150 marks in aggregate, securing at least 50 marks in the last paper. In how different ways can he secure marks in the three papers to qualify?
The above question is similar to finding all the non-negative integral solutions to the equation:
a + b + d = z
Where
a,b belong to [0,50]
d belongs to [50,100]
z belongs to [150,200]
Now how to solve after that....help required to get the answer. (The biggest problem is that z itself is variable)
Thanks in advance!!
UPDATE:
The answer is coming out to be a combo of three combinatorial series....need help after this:
       (102 C 100) + (103 C 101) + ..... + (152 C 150)

- 3×[ (51 C 49) + (52 C 50) + .....  +(101 C 99)] 

- 3×(51 C 48) =

 A: You can fix the 100 marks of the third exam. If you have 50 marks in the second exam, then you have 51 ways for the first exam. With 49 marks in the second exam there are 50 ways for the first exam. And so on. 
Therefore for the 100 marks of the third exam you have in total $51+50+49+...+1$ ways. This is equal to ${52 \choose 2}$.
For 99 marks of the third exam you have in total $50+49+...+1$ ways. This is equal to ${51 \choose 2}$.
And so on-until you reach 50 marks for the third exam.
In total you have $\sum_{k=2}^{52} {k \choose 2}$ ways. And this is equal to 
$\frac16\cdot (n+1)\cdot n\cdot (n-1)=\frac16\cdot 53\cdot 52\cdot 51=23,426$
Remark
$\sum_{k=m}^n {k \choose m}={{n+1} \choose {m+1}}$
In your case 
$\sum_{k=2}^{52} {k \choose 2}={{53} \choose 3}=\frac{53\cdot 52\cdot 51}{1\cdot 2\cdot 3}$
A: Taking $c=d-50$ the question can be rephrased as: 
How many solutions are there for: $$a+b+c\geq100$$
under the extra condition that $a,b,c\in\{0,1,\dots,50\}$?

Taking $u=50-a$, $v=50-b$ and $w=50-c$ it can be rephrased as: 
How many solutions are there for:$$u+v+w\leq50\tag2$$
under the extra condition that $u,v,w\in\{0,1,\dots,50\}$?

Fortunately if $u,v,w$ are nonnegative integers then $(2)$ implies that $u,v,w\leq50$ so it can be rephrased now as: 
How many solutions are there for:$$u+v+w\leq50$$
under the extra condition that $u,v,w$ are nonnegative integers?

This can be rephrased as: 
How many solutions are there for:$$r+u+v+w=50$$under the extra condition that $r,u,v,w$ are nonnegative integers?

Final answer found with stars and bars:$$\binom{53}{3}=23426$$
