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Combinations
By Philip Ellison on Wednesday, January 02, 2002 - 01:42 pm:

How many ways are there of ordering m items into n spaces? I thought it might be m!/(n!((m-n)!))
but this suggests that, for example, the odds of winning the lottery would remain unchanged if we had to pick 43 numbers, and not 6.
Thanks for any help

By Yatir Halevi on Wednesday, January 02, 2002 - 02:38 pm:

Philip, that is exactly it.
The odds wont change at all!
Lets say we have 5 numbers to choose from and we have to choose 2.
5!/(2!3!)=10
10 possibilities:
(1,2)
(1,3)
(1,4)
(1,5)
(2,3)
(2,4)
(2,5)
(3,4)
(3,5)
(4,5)

Picking 3 out of 5, we get:
5!/(3!2!)=10
the ten possebilities are:
(1,2,3)
(1,2,4)
(1,2,5)
(1,3,4)
(1,3,5)
(1,4,5)
(2,3,4)
(2,3,5)
(2,4,5)
(3,4,5)

There for the odds of me picking 2 out 5, is 1/10. And the odds of me picking 3 out 5 is also 1/10.

Hope it helped clear it up.

Yatir

By Dan Goodman on Wednesday, January 02, 2002 - 04:40 pm:

Actually, what you've worked out is the number of ways of choosing m from n, but not ordering m from n. This is what is going on in the lottery.

Another way of looking at why the number of ways of choosing m from n is the same as the number of ways of choosing n-m from n is that if you choose m things, then there are n-m things you haven't chosen, and these n-m things are completely determined by the m things you did choose. In other words, for each choice of m things there is a unique choice of n-m things that corresponds to it. So the number of choices of m things is the same as the number of choices of n-m things.

By Yatir Halevi on Thursday, January 03, 2002 - 12:01 pm:

Dan,
Are you sure that in the lottery we order m from n, because the order of the elements is not important, only the numbers that we get...

Yatir

By Dan Goodman on Friday, January 04, 2002 - 02:22 am:

Yatir, you're right. That's what I meant in my previous message, but I agree I didn't word it very well. In the lottery we're choosing, not ordering.