First we show that we can divide the five points into sets $S$ and $T$ according to the second condition in only one way. Assume that we can divide the five points into $S \cup T$ and $S^{\prime} \cup T^{\prime}$. Then, let $A=S^{\prime} \cap S, B=S^{\prime} \cap T, C=T^{\prime} \cap S$, and $D=T^{\prime} \cap T$. Since $S, T$ and $S^{\prime}, T^{\prime}$ partition the set of five points, $A, B, C, D$ also partition the set of five points. Now, according to the second condition, there can only be segments between $S$ and $T$ and between $S^{\prime}$ and $T^{\prime}$. Therefore, the only possible segments are between points in $A$ and $D$, or between points in $B$ and $C$. Since, according to the first condition, the points are all connected via segments, it must be that $A=D=\varnothing$ or $B=C=\varnothing$. If $A=D=\varnothing$, then it follows that $S^{\prime}=T$ and $T^{\prime}=S$. Otherwise, if $B=C=\varnothing$, then $S^{\prime}=S$ and $T^{\prime}=T$. In either case, $S, T$ and $S^{\prime}, T^{\prime}$ are the same partition of the five points, as desired. We now determine the possible sets of segments with regard to the sets $S$ and $T$. Case 1: the two sets contain 4 points and 1 point. Then, there are $\binom{5}{1}=5$ ways to partition the points in this manner. Moreover, the 1 point (in its own set) must be connected to each of the other 4 points, and these are the only possible segments. Therefore, there is only 1 possible set of segments, which, combining with the 5 ways of choosing the sets, gives 5 possible sets of segments. Case 2: the two sets contain 3 points and 2 points. Then, there are $\binom{5}{2}=10$ ways to partition the points in this manner. Let $S$ be the set containing 3 points and $T$ the set containing 2 points. We consider the possible degrees of the points in $T$. - If both points have degree 3, then each point must connect to all points in $S$, and the five points are connected via segments. So the number of possible sets of segments is 1. - If the points have degree 3 and 2. Then, we can swap the points in 2 ways, and, for the point with degree 2, we can choose the elements of $S$ it connects to in $\binom{3}{2}=3$ ways. In each case, the five points are guaranteed to be connected via segments. Hence 6 ways. - If the points have degree 3 and 1. Similarly, we can swap the points in 2 ways and connect the point with degree 1 to the elements of $S$ in $\binom{3}{1}=3$ ways. Since all five points are connected in all cases, we have 6 ways. - If both points have degree 2. Then, in order for the five points to be connected, the two points must connect to a common element of $S$. Call this common element $A$. Then, for the other two elements of $S$, each one must be connected to exactly one element of $T$. We can choose $A$ in 3 ways, and swap the correspondence between the other two elements of $S$ with the elements of $T$ in 2 ways. Hence 6 ways. - If the points have degree 2 and 1. Then, in order to cover $S$, the point with degree 2 must connect to 2 points in $S$, and the point with degree 1 to the remaining point in $S$. But then, the five points will not be connected via segments, an impossibility. - If both points have degree 1. Then, similar to the previous case, it is impossible to cover all the 3 points in $S$ with only 2 segments, a contradiction. Combining the subcases, we have $1+6+6+6=19$ possible sets of segments with regard to a partition. With 10 possible partitions, we have a total of $19 \cdot 10=190$ possible sets of segments. Finally, combining this number with the 5 possibilities from case 1, we have a total of $5+190=195$ possibilities, as desired.