December 18th, 2010, 07:26 PM  #1 
Newbie Joined: Dec 2010 Posts: 11 Thanks: 0  Counting problem
Hi I've encountered a counting problem. I think I've calculated the answer but I don't have the solution for checking. 49 balls marked 1, 2, 3, 4, ..., 48, 49 Pick 6 balls from them (49C6 combination, right?) Assume we have A1, A2, A3, A4, A5 and A6 picked from the 49 balls such that An  An1 >= 5. How many combination are there? First I define a bijection between picked balls and a binary string. 1, 2, 3, 4, 5, ..., 46, 47, 48, 49 For example, if ball 1 is picked, we then use 1 to represent ball 1, 2, 3, 4, 5. Similarly, if 46 is picked, we use 1 to represent ball 46, 47, 48, 49. For any balls that are not picked, we use a 0 to represent it. If ball 46 is one of the six balls, we get (49(5*5+4))C6 different combination. If ball 47 is one of the six balls, we get (49(5*5+3))C6 different combination. If ball 48 is one of the six balls, we get (49(5*5+2))C6 different combination. If ball 49 is one of the six balls, we get (49(5*5+1))C6 different combination. If none of 46, 47, 48 or 49 is selected, we get a general case (49(5*5+5))C6 different combination. So we sum them all up and get 295716 different combination. Are there any bugs in the calculation? Thanks in advance. 
December 20th, 2010, 08:44 PM  #2 
Newbie Joined: Sep 2009 Posts: 20 Thanks: 0  Re: Counting problem
I want to know about your bijection more formally. If you are familiar with programming, you can write an algorithm with the bijection and your way running in your brain. Then, the verification for your answer is easier. 

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