For those not familiar with Bridge, Whist, & similar card games all you need to know is that these games are played with a standard 52-card deck & 13 cards are dealt to each of 4 players. Some easy warmup questions. How many possible hands can a player hold? Answer: 52! / 39! * 13! = 635 013 559 600 How many deals to 4 players are possible? Answer: 52! / (13!)[sup]4[/sup] This is 5.36 * 10[sup]28[/sup] How many different holdings are there in a Particular suit? 2[sup]13[/sup] = 8192 For almost all practical purposes the following holdings in a particular suit are equivalent: AK432, AK753, AK642 When talking about a hand, players refer to such holdings as AKxxx. Considering some cards as equivalent brings to mind some more difficult problems. How many hands are possible if the 2 through the ten are considered to be equivalent? How many deals are possible? How many holdings in a particular suit? For the last question, the answer is 10*2[sup]4[/sup] = 160, which is obviously a lot less than 8192 The ten & nine are often important in the play of a hand. How many hands are possible if the 2 through the 9 are considered to be equivalent? How many deals? How many holdings in a suit? 9*2[sup]5[/sup] = 288 for the last question. Suppose the 2 through the 8 are considered to be equivalent. How many hands? How many deals? 8*2[sup]6[/sup] = 512 for the number of possible suit holdings. A program could be written to answer the above unanswered questions, but I know of no easy analytical method. Does anyone have answers to the above? I might take the time to write a program to answer these questions.