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TheShootah
04-12-2005, 01:39 AM
Can someone explain this choose business to me, like C(50,2) or something like that, with an example. I know a good deal about simple probability from AP Stats, but I was looking at a good inclusion - exclusion post and don't really know how to work out the C(50,2) Obviously it's 50 choose 2 and that is the probability of being dealt a specific hand or something, but I dunno really how to do it, unless it's just (2 / 52) * (1 / 51), which might work. I dunno. I have no idea. Thanks....

jason_t
04-12-2005, 02:38 AM
C(n,k) is the number of ways to choose k distinct objects from a set of n distinct objects and order doesn't matter. C(n,k) is known as the binomial coefficient because it appears in the expansion of the binomial (x+y)^n. It turns out that

http://i3.photobucket.com/albums/y73/jason_t712/formula1.jpg

which is useful in working with C(n,k). Why is this true? Here's one reason. (There are many.)

Pick the first object. There are n objects to choose from. Pick the second object. There are (n-1) objects to choose from. Pick the third object. There are (n-2) objects to choose from. And so on down to the kth choice of which there are (n-k+1) objects to choose from. There are n(n-1)(n-2)...(n-k+1) possible ways to do this. However, order does not matter, so we have to divide this result by the number of ways to order k objects. How many ways are there to order k objects? There are k choices for the first object, and (k-1) choices for the second object and so on. Therefore, there are k(k-1)(k-2)...1 ways to order k objects. Hence,

http://i3.photobucket.com/albums/y73/jason_t712/formula2.jpg

which isn't that pretty looking. But what does that first formula reduce to? This is simple; it's just

http://i3.photobucket.com/albums/y73/jason_t712/formula3.jpg

and we are done.

So, in particular, C(50,2) = 50!/(2!(50-2)!) = 50!/(2!48!) = 50 * 49 / 2 = 1225. That's the number of ways of choosing two cards from fifty. It's also the number of ways of choosing two pizza topping from fifty, and so on. That's the specific example that you asked for.

If you think about this, and try following the math, it should help. Please let me know if it doesn't.

Edit: Two typos fixed as a result of niin's reply. Thanks!

niin
04-12-2005, 02:43 AM
Combinations are the number of possible arrangements of r things out of a set of n items, when order doesn't matter. It's the permutations of r out of n items, divided by the total way that r things can be arranged (so that XY and YX aren't counted as two distinct selections). The formula is:

<font class="small">Code:</font><hr /><pre> n!
C(n,r) = ---------
(n-r)!r!
</pre><hr />

EDIT: Jason's post is longer. Jason, question tho, shouldn't it be as I list it here? Your formula reduces to 1/(n-r)! .

TheShootah
04-12-2005, 10:46 AM
Thanks guys, not too shabby! /images/graemlins/grin.gif

EDIT: Let's keep going to see if I really get it. The reason I wanted to learn this was to be able to do some calculations about poker (duh). This might be useful to figure out how to calculate the probability of you being dealt Aces. How would I do that?

closer2313
04-12-2005, 11:30 AM
To figure out the probability of being dealt pocket aces you could do it a couple of ways.

First way
Probability of being dealt an ace in the first card * probability of being dealt an ace for the second card.

(4/52) * (3/51) = .0045 = 1 / 221 times

Doing it with combinations.

4 aces choose 2 / 52 cards choose 2

4 choose 2 / 52 choose 2 = 0.00452488688 = 1 / 221 times

TheShootah
04-12-2005, 12:19 PM
Got it...Thanks!