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If you're studying A-level statistics, you will have come across the equation on the right. The bracketed expression on the left hand side is pronounced "n choose k", and it's the number of different ways of choosing k out of n things. For example, in the UK National Lottery, there are 59 numbered balls, and six are drawn every week (we'll ignore the bonus ball). How many different ways are there of choosing six balls out of 59? A mathematician will answer "59 choose 6". But what does that come to?
Let's take it step by step. There are 59 possible choices for the first ball, and then 58 for the second, and so on, until finally there are 54 choices for the sixth ball. So it looks as if the answer might be 59 x 58 x 57 x 56 x 55 x 54, which comes to 2,147,483,647, or a little over two billion. Remember, the factorial of a number (written with "!" after it) is the number itself, times one less, times two less...all the way down to 1. So 59! is in fact 59 x 58 x 57 x 56 x 55 x 54 x 53!, and therefore our just-over-two-billion value is in fact 59!/53!.
But wait: for lottery purposes, we don't actually care which order those six balls were drawn in. Drawing 23, 12, 15, 7, 42, and 19 is essentially the same as drawing 12, 7, 42, 19, 23 and 15, for example. How many possible orderings are there for six balls? We have six choices for the first ball, five for the second, and so on, so the answer is 6!, which is 6 x 5 x 4 x 3 x 2 x 1 = 720. So our two billion different draws actually fall into groups of 720, where the identity of the balls (though not their ordering) is the same within each group. And it's the number of groups we care about. So we divide our big number by 720, and get 45,057,474. That's the value of 59 choose 6: 59! / (53! x 6!).
And one divided by that number is your chance of winning the jackpot with any given set of six numbers. Lottery tickets cost £2 each, and I will leave it to you to work out whether they're a good investment (you can probably just use common sense rather any heavy maths).
Tip: if k is small, you can work out "n choose k" on paper this way. As the numerator of a fraction, write the value of n, then n-1, then n-2, until you've got r numbers in total. For the denominator, write k, k-1 down to 1. So you have k different numbers on both top and bottom. For example, 10 choose 3 is just 10 x 9 x 8 / 3 x 2 x 1. You can divide the 3 on the bottom into the 9 on the top, and the 2 on the bottom into the 8 on the top, leaving you with 10 x 3 x 2 = 60. As special cases of this, "n choose 1" is always n, while "n choose 0" is always 1.
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