Probability of picking one card without another one. I'm having some troubles understanding a probability question. 
In a game, there 21 assets and we pick 5 of them. 
Let the events : 
A = "pick the first asset" 
B = "pick the last asset"
We are ask to find p = P("A or B")
I've said that "A or B" = "A and B" + "A without B" + "B without A"
So first I can compute P("A and B ") = nCr(2,2)*nCr(19,3)/nCr(21,5)
where nCr(n,k) represents 'n choose k'. This part is correct yet.
But I don't understand why the probability to pick one card without the other one isn't given by : P("A without B") = nCr(2,1)*nCr(19,4)/nCr(21,5)
In the correction, the resolution goes another way : 
it says that "A or B" = A + B - "A & B", and then compute P(A) = P(B) 
I understood why that was correct, but I cannot understand where my mistake is and I would appreciate very much some help here. 
Thank you,
Arthur

For pick one without the other, the $nCr(2,1)$ is the number of ways to choose one of the two and the $nCr(19,4)$ is the number of ways to choose four cards from the remaining $19$.  This will give the number of ways to get exactly one of A or B.  If you want A without B you should change the $nCr(2,1)$ to $nCr(1,1)$ because you are only selecting from A for the first card.
