Consider the question: what is the probability of getting a 6-card poker hand that contains a card from all four suits (order doesn't matter):
Each of the C(52, 6) hands is equally likely. Let E be the event that the hand selected contains 6 a card from all four suits. Then
Pr(E) = |E| / C(52, 6)
For |E|, first pick one heart, then one spade, then one diamond, then one clubs, and, finally, two additional cards. Therefore
|E| = 134 . C(48, 2) and hence Pr(E) = (134 . C(48, 2)) / C(52, 6)
Explain what is wrong with this solution. If there is over-counting in |E|, characterize all hands that are counted more than once, and how many times each such hand is counted.
What is the correct answer for Pr(E)?