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### Winning the UK National Lottery

```
Date: 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!

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

```
Associated Topics:
High School Permutations and Combinations

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