Winning the UK National LotteryDate: 11/8/95 at 19:8:5 From: Dominic Parkinson Subject: Lottery Thank you for a superb reply recently regarding Infinity. I now need to know the odds on the UK National Lottery: punters choose 6 DIFFERENT numbers, each between 1 & 49. How many permutations? 49! - 43! gives 10 billion, which is wrong since the order of the numbers chosen is not relevant. How can I calculate the reduced number of permutations to reflect the fact that the order of the chosen numbers is not relevant? May you win a lottery if you can tell me. (Do I divide by 6! ?) Dominic. Date: 11/12/95 at 14:45:50 From: Doctor Jeremy Subject: Re: Lottery What you need to do is take the number of ways to choose 6 different numbers, and divide that by the number of such ways are identical. For the first part, I assume when you wrote 49!-43! you meant 49!/43!; the latter is a lot closer to 10 billion than the former (which is a lot larger). For each permutation, there are not 6, but 6! different orders for it to appear in (it's all the orders in which we can choose 6 out of those 6). So the result is 49! --- 43! * 6! or about 14 million. The way to think about this is that each selection of 6 elements can be reordered 6! ways. So for each of the 49!/43! ways of selecting 6 elements, it is one of 6! ways that are, as far as we care, identical. -Doctor Jeremy, The Geometry Forum |
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