Suppose you have a deck of $20$ cards with $4$ suits and ranks {A, K, Q, J, 10}. Let's say you draw 4 cards at random (without replacement). What is the probability you will have exactly 3 suits?
I tried $\frac{20\cdot4\cdot15\cdot10}{20\cdot19\cdot18\cdot17}$ but this is not correct I think. What am I doing wrong and what would instead get me the correct answer?
This problem is easier if we consider it without order.
Number of possible 4-card hands : ${20 \choose 4}=4845$
Number of hands with exactly three suits : ${4\choose 1}{5\choose 2}{3 \choose 2}{5\choose 1}{5\choose 1}=3000$
Probability $\frac{3000}{4845}=\frac{200}{323}$