In how many ways can a student score exactly $100$ points on four $50$ point exams? greater than $100$? A student takes up $4$ exams with $50$ points each. In how many ways can he score exactly $100$? Similarly, in how many ways can he score greater than $100$?
My try:
I cannot seem to fill in the $4$ possibilities. The range depends on the constraint of equating to $100$.
Thanks for any help!
 A: To score exactly 100, first suppose there are no constraints other than having nonnegative numbers. Then the number of solutions to $a+b+c+d = 100$ is simply $\binom{103}{3}$ by a standard stars and bars argument. Now we need only remove solutions where one of the values is greater than $50.$ Suppose $ a > 50,$ then we ask for solutions to $(a-51) + b + c + d = 49,$ where $a,b,c,d$ are nonnegative, which has $\binom{52}{3}$ solutions. Since we could have had any of the four be the one larger than $50,$ there are $4\cdot \binom{52}{3}$ such "bad" solutions. Hence the required total is $N_{100} = \binom{103}{3} - 4\binom{52}{3}.$
For the second part, any solution to $a + b + c +d > 100$ corresponds with $(50-a) + (50-b) + (50-c) + (50-d) < 100.$ Hence the number of such solutions is just $(51^4 - N_{100})/2,$ where $N_{100}$ is the answer to the first part
A: Here is one brute-force approach which is less elegant than the other answer but perhaps will showcase a different way of arriving at the same number.
Let's split up our exams into two groups. Exam one and two are in group $A$ and exam three and four are in group $B$. Let $50 \leq n \leq 100$. We will want to consider the number of ways that $n$ points are obtained in group $A$ and the remaining $(100 - n)$ in group $B$.
First, suppose $n = 50$. The number of ways for group $A$ to score $50$ is $51$, since there is one possibility for each outcome of exam one, ranging from $0$ to $50$. There are a similar number of possibilities for group $B$. Hence $(51)^2$ total possibilities exist in this case.
Otherwise, one group scores above $50$ and the other below $50$. Suppose that it is group $A$ that scores higher -- we can justify this assumption by multiplying by $2$ at the end, as there are identical cases when group $B$ scores higher. For $n>50$ there are $(101 - n)$ ways for group $A$ to score $n$ points. For example if $n = 100$ then there is one way, and if $n = 60$ there are $41$ possibilities, as exam one can receive between $10$ and $50$ points. Next, with the remaining $(100-n)$ points for group B there are $(100-n)+1 = 101-n$ possibilities. This is because, for example, if $n=60$ then $100-n = 40$ and there are $41$ possibilities, as exam $3$ can score between $0$ and $40$ points. Thus the total number of possibilities are $(101 -n)^2$. This formula is nice and symmetric, and suggests that a more beautiful analysis than the one here is possible. Anyhow, we get the sum total number of possibilities as 
$$(51)^2 + 2\sum_{n=51}^{100} (101 - n)^2 = (51)^2 + 2\sum_{n=1}^{50} n^2$$
Using a simple formula for the sum of squares, we get $(51)^2 + 2(42925) = 88451$.
So far so good. How do we compute the total number of ways of score above a hundred? By symmetry, this is equal to the total number of ways of getting less than $100$, simply by replacing every right answer by a wrong answer. So the total number of possible scores $(51)^4$ is the sum of $88451$ and twice the score in question. We compute
$$ \frac{51^4 - 88451}{2} = 3338375$$.
A: The scores he can get  are the coefficients of the different powers of $x$ in the expansion
\begin{align*}
(1+x+\cdots + x^{50})^4 &= \left(\frac{1-x^{51}}{1-x}\right)^4 \\
&= (1-4x^{51} + 6X^{102} - 4X^{153} + X^{204})\left(1+ 4x + \frac{4\cdot 5}{1\cdot 2}x^2 + \frac{4\cdot 5 \cdot 6}{3!}x^3 + \cdots +\binom{n+3}{3}x^n+\cdots \right)
\end{align*}
Hence the number of ways in which he can get exactly 100 is 
$$ \binom{103}{3} - 4 \binom{52}{3} = 88451$$
Second part has already been solved in the other solutions posted.
A: Each score goes from 0 to 50 so we can decompose it this way: $P(a+b+c+d = 100) = \sum\limits_{X=0}^{100}{P(a+b = X)P(c+d=100-X)}$
$P(a+b = X)=\frac{1}{51^2}card\{(0,X), (1,X-1), ..., (X,0)\} = \frac{X+1}{51^2}$
$P(a+b+c+d = 100) = \sum\limits_{X=0}^{100}(\frac{X+1}{51^2}\frac{100-X+1}{51^2})$
$=\frac{1}{51^4}\sum\limits_{X=1}^{101}X(102-X)= \frac{1}{51^4}\sum\limits_{X=1}^{101}(102X-X^2)$
$=\frac{1}{51^4}(-\frac{(101.(101+1).(2.101+1)}{6}+102.\frac{101.(101+1)}{2})= \frac{176851}{6765201}\approx0.0261$
Edit: I just realized that you're asking for how many ways he can score 100 and not the probability of scoring 100. The answer you're looking for is the numerator of the above formula, so 176851.
