Date: 2 Aug 1995 18:59:46 -0400 From: Anonymous Subject: Bingo Question: Is there any literature on the number of ways of winning at bingo? Assume horizontal, vertical, and diagonal 5 in a row are required to win. How many possible different bingo cards are there?
Date: 3 Aug 1995 11:26:08 -0400 From: Dr. Ken Subject: Re: bingo Hello there! One of the Doctors, Heather Mateyak, knew about the configuration of Bingo cards, so she went ahead and solved the problem. Here it is. The first row can contain the numbers 1-15 in any order, with no duplicates. The second number can contain 16-30, the third 31-45, the fourth 46-60 and the fifth 61-75. There's also a free space in the center square that doesn't get a number. So how can we fill in the spaces, given that these are our choices for filling them in? Well, there are fifteen ways of filling in the first square, fourteen ways of filling in the next one down (since we already chose the first square and can't duplicate the number), thirteen for the next, then twelve, then 11. So there are 15*14*13*12*11 different versions of the first column. Likewise, there are the same number of different versions of every other column except the middle one, for which there are 15*14*13*12 different versions. So to get the total number of different possible Bingo cards, we multiply all these together, and we get (15*14*13*12)^5 * 11^4, which is about 5.5 x 10^26. -K
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