Tea Party with Liars

Knights always tell the truth, so the placement for them is fixed (must have a knave (noted as k) on their left and a Joker (noted as J) on their right). So between 2K, 3k & 4J, we have to set aside 6 of them to accommodate the Knights with two groups of kKJ (say we arranged them from behind).

(kKJ) + (kKJ) left behind a knave (1k) and 2 Jokers (2J). Since Jokers can both lie and tell the truth, we only need to care if the knaves can stick with their lies. Since all of them are saying the same thing and the statements have an AND connective, we just need one of the statements to be false for a knave to be able to say it.

The knaves in a group don't have a Joker on their right so it shouldn't matter if they have a fellow knave on their right. On the other hand, the only solitary knave can't have another knave on their left (all the others are already in groups and none left as a fellow soloist) so it doesn't matter if they (the solitary one, gender unknown) have a Joker on their right.

So, now we know that we have two indistinguishable groups, two indistinguishable Jokers and one solitary knave that can be arranged in anyway in a straight line since it won't compromise with the falsity of the knaves involved.

Answer
= n(number of arrangements of 5 parties) / n(number of indistinguishable factor)
= 5! / 2!2!
= 120 / (2 × 2)
= 30

As the knight have to tell the truth, you know that it has to look like _ k K j _ k K j _ where the k is a knave, K is a knight, a j is a joker and a _ is a possible space for the last k or j's.

You can then put the last k is 3 places ( _ ) , not changing the knights' statements. if you start with _ k _ k K j _ k K j _ with the last knave at the start, there are four places for the joker to go.

On the first gap, there are still 4 place left as another joker can go beside them. If the first extra joker when in the second possible place, it would leave 3 places for the last one as if you put one in the first gap it would be the same as a solution in the first gap. If you carry on and check all the statements, you have 4+3+2+1=10 solution for the jokers.

If you go back to where the extra knave goes, there are 3 solutions, all providing 10 out comes for the jokers and 3*10=30