How many ways are there to arrange five cards in a row so that each arrangement begins or ends with a club? 
How many ways are there to arrange five cards in a row so that each
  arrangement 
(1) begins or ends with a club? (2) contains exactly two kings?

I have an idea of where to start with Question 1. There should be three cases.
Case 1: begins with a club: 4 * P(48,4)
Case 2: ends with a club: P(48,4) * 4
Case 3: overlap (begins and ends with a club): 4 * P(50,3) * 3 
Case 1 + case 2 - Case 3 = answer
I'm pretty stuck because I am NOT getting the answer I should be. 
For Question 2, I am not sure how to approach this. How would I deal with selecting these two kings if they can be put in any of the 5 spots available? 
 A: *

*There are 13 clubs in a deck, and you're only putting one in the end position. So your case 1 and 2 should be P(51,4) * 13. You've got 13 options for the club in the end position, then choosing four out of the remaining 51 cards to add. Similarly, for case 3 you'd want to change the 4 and 3 to 13 and 12 -- so 13 * P(50,3) * 12.
What you have now would be more applicable for a question along the lines of "How many ways can you arrange five cards so that there is an Ace in an end position".  

*There are five ways to place the first king, times four possible suits it could be. There are four ways to place the second king, times three possible suits (since we've already placed one). Then ignoring the other two kings, there are 48 cards left, and we want to arrange three of them - good old P(48,3).

A: (1) How many ways are there to pick a club for the first card? Given that one club has been picked already, how many ways are there to pick a club for the last card (which we will pick second, then slip the other three cards inbetween)? Given that two clubs have been picked already, how many ways are there to pick the remaining three cards that we don't care about? The product of these three numbers is the answer.
(2) how many ways are there to pick the first king? how many ways to pick the 2nd king? Now just throw out the other two kings. How many ways are there to pick the other 3 cards? The product of these three numbers is the answer.
A: For $(2)$, to get the answer of $12453120$,
Pick 2 kings and 3 non-kings and permute, i.e. $\dbinom{4}{2}\dbinom{48}{3}\times5!$
