If the Flamingo is chosen, we immediately also choose the Albatross and the Hummingbird. This leaves only two spots to be filled. We can eliminate the Eagle because it cannot appear with the Flamingo, and we can eliminate the Diver because it cannot appear with the Albatross. We are left to choose from the Bluebird, the Condor and the Gull. If we choose the Gull, we cannot also choose the Bluebird - since the Hummingbird is already present, this situation would violate the rule involving the Gull. Therefore, we must chose either the Gull OR the Bluebird, and then flll the remaining spot with the Condor. The two distinct possibilities are: Flamingo, Albatross, Hummingbird, Condor and Gull OR Flamingo, Albatross, Hummingbird, Condor and Bluebird.

Knowing that the Diver is chosen allows us to immediately eliminate the Albatross according to the first rule. Eliminating the Albatross causes us to also eliminate the Flamingo in accordance with the second rule (the contrapositive of this rule states that if either the Albatross or the Hummingbird is NOT chosen, the Flamingo is also not chosen). We know we can only choose two out of the Gull, the Hummingbird and the Bluebird because of the last rule. This leaves two final empty spots which must be filled by the Eagle and the Condor. 

When we eliminate the Hummingbird we must also immediately eliminate the Flamingo. This leaves us with six leftover variables to choose from. Since the Diver and the Albatross can never be chosen together, we know that the four other variables, namely the Gull, the Blurbird, the Condor and the Eagle all must be chosen. The only option is whether to choose the Albatross or the Diver to complete the group.

We are trying to figure out if it is possible for four or more students to have black hair, since we know there are exactly three students with brown hair. Since three of the eight students have brown hair, that leaves five. Based on the second and fifth condition we know one of those five has red hair, which leaves four. Based on the fifth conditon we know there must be at least one blonde, which leaves three. Therefore, the absolute maximum number of students with black hair would have to be three, which is not more than the number of students with brown hair.

We know that of the eight students, three of those students have brown hair, which leaves five. Based on the second and fifth conditions, we know that one of those students has red hair, which leaves four. Based on the fifth condition, we know that at least one of those students has black hair, which leaves three. Therefore, the absolute maximum number of students with blonde hair is three.

There is at least one possible scenario in which no girl has the same colored hair as any of the boys in the class. It is possible, for instance, that there are only two boys in the classroom. In this case, based on the third condition we know that the boys will have the same color hair. If the two boys have black hair, then that leaves six girls. Based on the first condition, we know that four girls have either blonde or brown hair, neither of which is black. That leaves two girls. Based on the second and fifth condition we know that one of those girls has red hair, which leaves one girl. It is not specified that ONLY four girls have brown or blonde hair, which means that more than four girls could have either blonde or brown hair. If the final girl has blonde or brown hair, then no girl has black hair, and therefore no girl has the same colored hair as any of the boys in the classroom.

Based on the first condition, we know that four girls have either blonde or brown hair. Based on the second condition and the condition posed within this answer choice, we are assuming that the one student with red hair is a girl. This means that there are three students left. Based on the condition posed within this question, we know that those three students must be boys. Based on the fifth premise, we know that at least one of each hair color must be represented. Since black hair is not represented by any of the girl students, we know that at least one of the boys must have black hair.

This is the only scenario that meets all of the conditions: four girls have either blonde or brown hair (three have blonde hair and one has brown), only one student has red hair, two boys have the same hair color (brown), there are exactly three students with brown hair (two boys and one girl), and at least one of every hair color is represented.

Dragon: Q, T, V

Salamander: R

Phoenix: S

Rabbit: P, U

This one breaks the rule that QT cannot be paired together if RV are not paired together. This is more clear if you write out the contrapositive.

~(RV) --> ~(QT)

(QT) --> (RV)

Dragon: V, P

Salamander: S

Phoenix: Q, R

Rabbit: T, U

This is incorrect, because it breaks the rule that Dragon must have more new recruits than any other team, not simply as many. This incorrect answer, if properly considered, also leads to the key insight that Dragon will always have at least three new recruits. That is because the distributions of new members, in order to satisfy both rules that Dragon must have the most and each team must have at least one, will look like either 4/1/1/1 or 3/2/1/1. There are no other possible distributions. Furthermore, if it is a 3/2/1/1 distribution and P is NOT a Dragon, then P is in the team with two new recruits, since P must be in a team with more new recruits than S (who must be a team with only one new recruit).

Dragon: R, V, Q, T

Salamander: U

Phoenix: P

Rabbit: S

This is incorrect, because S and P are on teams that have the same number of new recruits. P must be on a team with more. 

Dragon: P, V, U, Q

Salamander: T

Phoenix: R

Rabbit: S

This is incorrect because T is not on Dragon, but U is on Dragon. This is another one that becomes more clear if you write out the contrapositive.

~(Td) --> ~(Ud)

Ud --> Td

[Q is assigned to Dragon and U is assigned to Dragon.]

If U is assigned to Dragon, then T must also be assigned to Dragon. This is the contrapositive of the rule that U is not in Dragon if T is not in Dragon.

Dragon: Q, U, T

Salamander:

Phoenix:

Rabbit:

If Q and T are together, then R and V must also be together. This is the contrapositive of the rule that if R and V are not together, then Q and T are not together. Now, we can't put both R and V in Dragon, because then there wouldn't be enough recruits to distribute to other teams. For now, let's put them in Salamander.

Dragon: Q, U, T

Salamander: R, V

Phoenix:

Rabbit:

We must still deal with P and S. P must be in a group with more new recruits than S, so it it would have to go in Dragon or Salamander. Unfortunately, that would mean that either Phoenix or Rabbit would be left without a recruit, which is unacceptable.

Thus, we know that Q and U cannot both be in Dragon together.