Each incorrect answer violates at least one of the conditions.

The third condition eliminates {A, B, C, L}. All three beef can't be on the menu.

The fourth condition eliminates {A, B, L, N}. A and N cannot be on the menu together.

The last condition eliminates {A, B, L, M} and {B, L, M, N}. If M is on the menu, C and X must also be on the menu. In each case, M is on the menu without either C or X and with only one dish left to be chosen, which makes fulfillment of the condition impossible.

{A, X} and {A, Y} can be eliminated through a combination of the first and second conditions. If A is on the menu, then Z must be on the menu (second condition). That means that neither X nor Y can also be on the menu (first condition).

{M, Y} and {M, Z} can be eliminated through a combination of the first and last conditions. If M is on the menu, then X must be on the menu (last condition). That means that neither Y nor Z can also be on the menu (first condition).

So, only the correct answer cannot be eliminated.

The key to answering this question is recognizing that the first condition (one fish) and the third condition (no more than two beef), taken together with the fact that the tasting menu always consists of five dishes, imply that there must be at least two poultry dishes on any menu.

Then, if Z is on the menu, neither X nor Y can be on the menu (per the first condition). Further, if X cannot be on the menu, neither can M (per the last condition). 

Finally, since there must be at least two poultry dishes on any menu and M cannot be one of those poultry dishes, L and N (the remaining poultry dishes) must both be always be on any menu containing Z.

The correct answer is the only one that does not violate any of the conditions.

If A is chosen, Z must be chosen (second condition), but then X cannot be chosen (first condition). So, {A, L, X} is eliminated.

A and N cannot be chosen together (fifth condition), so {A, N, Z} is eliminated.

If M is chosen, X must also be chosen (last condition), but then Y cannot also be chosen (first condition). So, {B, M, Y} and {C, M, Y} are eliminated.

The correct answer is the only one that does not violate any conditions.

{A, B, X, Y, Z} has B without D, so it violates Condition 4.

{A, C, D, W, X} has C but not E, so it violates Condition 5.

{A, W, X, Y, Z} contains only 1 Director from the first group, so it violates Condition 1.

{C, E, W, X, Z} contains both E and W, so it violates Condition 6.

All incorrect answers violate at least one condition.

X: H

Y: F, G, J, L

Z: K, N, O

cannot be right because only one attendant is watching X (Condition 1).

X: H, N

Y: F, K, L

Z: G, J, O

cannot be right because K is watching Y (Condition 3).

X: H, J, K

Y: G, N

Z: F, L, O

cannot be right because J and K are watching the same level, X (Condition 4).

X: G, H, O

Y: F, J, L

Z: K, N

cannot be right because G is watching X, while O is not watching Z (Condition 5).

This is a simple elimination by rule problem. First we can check for Amy and Ona, eliminating any answers that do not have them performing solo. Then we check for the conditional involving Carrie and Belle - any answer that has Carrie performing second and doesn't have Belle performing fourth is eliminated. Then we can check to make sure all remaining answers have only one performer in the first and last slots (they all do). Then we make sure Nicole and Linda are always performing together. Finally, any leftover answers are eliminated based on the rule that Ona must perform sometime before Dana.

This set up gives us three distinct possibilities. When we place Amy first and Carries second, we automatically add Belle fourth due to the conditional. Since Ona needs to perform alone and also needs to perform before Dana, the only spot left for her is third. Now we get into the possibilities. If Dana performs fourth with Belle, then we place Monique fifth and pair Nicole and Linda with Carrie. This yields the complete set list: First: Amy; Second: Nicole, Linda, Carrie; Third: Ona; Fourth: Dana, Belle; Fifth: Monique. The other option is to place Dana in the fifth number alone. This leaves Nicole, Linda and Monqiue. Since Nicole and Linda must be together, we have the option of placing them either with Carrie second or wth Belle fourth. Monique will be paried with the other. This yields two options, the first being: First: Amy; Second: Carrie, Nicole, Linda; Third: Ona; Fourth: Belle, Monique; Fifth: Dana. The second being: First: Amy; Second: Carrie, Monique; Thrid: Ona; Fourth: Belle, Nicole, Linda; Fifth: Dana. We can look through these possible templates and eliminate any answers that we never see in these hypotheticals.

If Belle performs first we know a few things right away. Since the first performance must be a solo, no one can perform with her. Therefore Ona cannot perform first - the earliest she can perform is second. Since Ona must perform before Dana, Dana cannot perform second. Based on the contrapositive of our conditional, we know that if Belle does not perform fourth, Carrie cannot perform second. And lastly, if Monique performs fifth we know that she must perform by herself. This would mean Amy and Ona would need to be solos in other spots, leaving four dancers to perform at once, which is not possible. 

If Ona performs second, we know that Dana must perform fifth. Ona must always perform solo, and the fifth spot must always be a solo so these two have to perform alone. Amy must perform solo, so she must go first in the other solo spot. Between spots two and three we must place Carrie, Belle, Nicole, Monique,  and Linda. If Carrie goes second, Belle must go fourth, which we know cannot happen in this scenario. Therefore Carrie must perform third. Nicole and Linda need to perform together, which leaves Belle and Monique to perform together. These two groups can switch between spots two and three. The two distinct possibilities for this scenario are: First: Amy; Second: Nicole, Linda; Third: Carrie, Belle, Monique; Fourth: Ona; Fifth: Dana OR First: Amy; Second: Belle, Monique; Third: Carrie, Nicole, Linda; Fourth: Ona; Fifth: Dana.