<s>[INST] 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!
 [/INST] 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
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