counting - pick 3 cards from a 52 card deck with restrictions I'm a little stuck on this problem... how do I get the cases/counts correct?
How many ways are there to pick three cards from a standard 52 card deck such that the first card is a King, the second card is not a diamond, and the third card is a Queen?
2 cases for first card:
King of diamonds
Not king of diamonds
If the first card is the king of diamonds, I have to make sure the king of diamonds doesn't come up as the 2nd card, and then need to make sure that the third card is a queen, but avoid over counting in case the 2nd card is a queen.
If the first card is not the king of diamonds, then I need to make sure the second card doesn't over count by including the king from the first card, and doesn't over count by dbl counting a queen in the 3rd card if a queen came up as the 2nd card.
I understand the first 2 cases, but the queen case trips me up.
So for the first 2, I have:
1*c(13,1)*c(3,1)
and 
c(3,1)*c(12,1)c(3,1)
But now how do I cover the possibilities of picking the queen?
 A: The answer is $600$.
We will distinguish the cases when the king and the queen are diamonds or not.
-Neither king nor queen are diamond: $3\times3$ possibilities for these two cards and it reminds 37 non diamonds for the second card. So here we have $3\times3\times37=333$ possibilities.
-The king is a diamond but not the queen: $1\times3$ possibilities for these two cards and it remind 38 non diamonds for the second card. So here we have $1\times3\times38=114$ possibilities.
-The king is not a diamond but the queen is: $3\times1$ possibilities for these two cards and it reminds 38 non diamonds for the second card. So here we have $3\times1\times38=114$ possibilities.
-The king and the queen are both diamond: $1\times1$ possibility for these two cards and it remind the 39 non diamonds for the second card. So here we have $1\times1\times39=39$ possibilities.
In total: $333+114+114+39=600$.
A: The "cases" computation is more smooth if we deal first with the second  restriction.
Case 1: Second is non-diamond non-King non-Queen: $33$ choices for that, then $4$ choices for King, $4$ choices for Queen, total $(33)(4^2)=528$.
Case 2: Second is non-diamond King: $3$ choices for that, then $3$ for King, $4$ for Queen, total $36$.
Case 3: Second is non-diamond Queen: Similar analysis to Case 2, same $36$ count.
Total: $528+36+ 36=600$
