Number of ways to divide n identical objects into groups Question 1:
If a student can score a maximum of $100$ marks in three subjects p,c,m, then find the number of ways in which he can score a total of $190$ marks while getting at least $50$ in each subject.
Procedure: 
$$x_1 + x_2 +x_3 = 190$$
$x_1 \geq 50, x_2 \geq 50, x_3 \geq 50$
which implies 
$$x_1+x_2+x_3 = 40$$
$x_1 \geq 0, x_2 \geq 0, x_3 \geq 0$
Number of ways $= \binom{40 + 3 - 1}{3-1} = \binom{42}{2}$
I understand this is the right answer.
Question 2: If a student can score maximum of $100$ marks in three subjects p,c,m, then find the number of ways in which he can score a total of $230$ marks while getting at least $50$ in each subject.
The formula used above gives $\binom{82}{2}$.
I understand this is not the right answer.
I also do understand that the above formula comes from multinomial theorem. I wanted to check why the formula works for question 1 and doesn't work for question 2. 
 A: In the first, you can just subtract $50$ from each score and find the ways to sum three numbers (including $0$) up to $40$  The stars and bars calculation for this is $42 \choose 2$ as you say.  
The difference in the second is that you need $80$ more points but no one test can give you more that $50$.  $82 \choose 2$ is the number of ways to add up three numbers (including $0$) to get $80$ if any of them can give you the full $80$.  Now you need to find the number of ways to add up three numbers in the range $[0,50]$ to get $80$.  
I don't have a neat way to count the second.  If you get $0$ points on the first subject there are $21$ possibilities.  If you get $30$ points on the first subject there are $51$ possibilities.  If you get $50$ points on the first there are $31$ possibilities.  It is linear between these, so the number of possibilities is
$$\sum_{i=0}^{30}(21+i)+\sum_{j=31}^{50}(81-j)=21\cdot31+\frac 12\cdot 30 \cdot 31+81\cdot 20-\frac 12\cdot 81\cdot 20=1926$$
A: For Question 1:
You can instantly ignore cases where any of the three scores are less than 50.  So assume you're getting at least 50 in each test.  Then the additional score needed is $40$ ($= 190 - 3 \cdot 50$).  Sum over these conditions with the total being $40$:
$$\sum\limits_{p=0}^{40} \sum\limits_{c=0}^{40-p} \sum\limits_{m=0}^{40-c-p} 1 = 12341$$
I'll let you apply the same logic to Question 2.
