4 Responses

1. a²+b²=c²

   November 20th, 2009 at 8:43 pm

   1

   There are 4 suits, and for each suit, 13 cards. You’re choosing 5 of them, so 4(13C5) = 5148 hands.
   You can also solve this by looking up the frequency of the types of hands in poker. What you want is the number of flushes plus straight flushes plus royal flushes.

2. icemetal

   November 20th, 2009 at 8:55 pm

   2

   Think of it this way: There are 13 cards of a single suit. So find the number of combinations of 13 cards, no repetition. Then realize there are 4 suits, so multiply that number by 4.
   The formula for combinations is n!/(r!(n-r)!) where n=number of possible cards (13) and r=number of chosen cards (5). Substituting gives us:
   13!/(5!(13-5)!) = 13!/(120*8!)
   If we work this out, we get 1287. Now we remember there are 4 suits, and multiply this by 4 to get:
   5148 combinations.
   -IMP

3. Bob K

   November 20th, 2009 at 10:14 pm

   3

   First, you choose which suit you want to have 5 cards of, so that would be 4C1, next, there are 13 cards in a suit and you want to choose 5 of those, or 13C5. And then multiply the two choices as they are independent
   4C1*13C5
   4!/3!1! * 13!/5!8!
   4 * 1287
   5148

4. kro

   November 21st, 2009 at 2:02 am

   4

   5,108 it’s called a flush.
   36 straight flush
   4 royal flush