Thinking about Possibilities!

First note that the scores of the five members must total to $5*(1 + 2 + 3 + 4 + 5) = 75.$ So since we are given that Andrew scores $24$ points, the scores of the other four must total to $51.$ Now since we are given that Eugene has $5 + 3 = 8$ points from two of the events, the minimum total score he can get is $11$ points. Since there are no ties, this means that the respective minimums for Daniel, Calvin and Brian are $12,13$ and $14,$ respectively. Since $11 + 12 + 13 + 14 = 50,$ there is one point left to distribute. If this point were distributed to anyone other than Brian then the total score of at least one of the others would have to be increased by one as well, which would make the sum of all scores exceed $75.$ Thus we know that the final scores of Eugene, Daniel, Calvin and Brian must be $11, 12, 13$ and $15,$ respectively, (if, in fact, there ends up being any solution to this problem).

Now since Eugene scores $5$ points in Number Theory, and since Andrew scores $24$ points in total, we can conclude that Andrew's scores in the five events must be $(5,4,5,5,5),$ (in the order that the events are listed). Also, since Eugene already has $8$ points but only receives $11$ points in total, his scores in the five events must be $(1,5,3,1,1).$

Next, for Calvin to receive the same scores in four events and to receive $13$ points in total, he must receive $3$ points in four events and $1$ point in the fifth. (This is because he cannot receive $4$ points in four events as his final score would then exceed $13,$ and he cannot receive $2$ points in four events as he would then need to receive $5$ points in the fifth, and all the $5$ point scores have already been distributed.) Now since Eugene receives $3$ points in Geometry, this implies that Calvin's scores in the five events must be $(3,3,1,3,3).$

Now since Daniel receives $4$ points in Algebra, Brian must receive $2$ points in this subject, as this will be the only score in this subject left to distribute. For Brian to reach a total of $15$ points, he must then receive $13$ points in the other four subjects. If he were to then receive only two of the remaining $4$ -point scores left to distribute, then he would have to receive $5$ points in the remaining two subjects. But since all the $3$ -point scores have already been distributed, there would be no way for two scores to add to $5.$ Thus Brian must receive $4$ points in three subjects and $1$ point in a fifth subject. Given that $4$ -point scores have already been awarded in Algebra and Number Theory, we can conclude that Brian's scores in the five events are $(2,1,4,4,4).$

This forces Daniel to have a score sheet of $(4,2,2,2,2),$ which does add to the required $12$ points. We have thus established a unique solution, and hence the answer is $\boxed{B}.$