What does choose mean and/or c(50,3)

[Jon Poker]

I am new to statistics. What does c(50,2) mean?

Also, how does the following combin(50,3) become 19600. What equation is used to turn (50,3) to 19600?
=4*(4*4*4)/combin(50,3)=256/19600=1.3%

Any help would be greatly appreciated.

[Cobra]

The term c(50,3) or combin(50,3), or (50 choose 3) is the total number of three card combinations you could have using 50 cards. A combination is a group of three cards were order does not matter. That is AKQ of diamonds is the same as KQA of diamonds, etc.

You can get the answer in a program like excel ( that is the combin equation) otherwise c(50,3) is equal to

50!/(47!*3!) were ! means factorial. This simplifies to
50*49*48/3*2*1. If you wanted c(50,2) that would be
50*49/2*1.

I believe there was recently a more detailed discussion of this in the probability area.

Cobra

[maryfield48]

User Zapped posted an excellent reply in that thread that I stored for reference, his post is quoted below, the complete thread is here:

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At a high level (24 choose 3) means how many ways can you pick three items out of a group of 24 items, if the resulting order doesn't matter and if we don't replace the items as we pick them.
In 2+2 context it's the total number of _different_ hands in a deal of 3 cards from a deck of only 24.

How do we calculate (24 choose 3)? Well, that requires a brief digression into the world of permutations & combinations. In permutations, order is important, but in combinations like (24 choose 3), order is not important. The permutation of 5[img]/images/graemlins/heart.gif[/img] then 2[img]/images/graemlins/club.gif[/img] then K[img]/images/graemlins/spade.gif[/img] would be considered a different result than if you got the same cards in a different order. Clearly 5[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/club.gif[/img] K[img]/images/graemlins/spade.gif[/img] is not a different hand than 2[img]/images/graemlins/club.gif[/img] K[img]/images/graemlins/spade.gif[/img] 5[img]/images/graemlins/heart.gif[/img] in poker, so we're usually not all that interested in permutations, except where they help us understand combinations

The "choose" in (24 choose 3) is shortand for a combination.
Intuitively you can imagine that there are fewer combinations (order unimportant) than permutations (order matters), but the key is how to calculate how many fewer combinations there are than permutations.

Calculating permutations is useful as a starting point to understand combinations. How many ways can you choose the first card from a deck of 24? Of course, 24 different ways. Now 23 cards remain, so there are 23 ways to choose the 2nd card, for a total of 24 * 23 = 552 different permutations. Now what if we said the order doesn't matter? Well, every permutation like 7 3 can be ignored if we already listed 3 7 .

In other words, the list of permutations can be cut in half for the list of combinations in this case. So if (24 permute 2) is 552, (24 choose 2) is [(24 permute 2) / 2 ] or 276.

What happens with three cards? You already probably realize that (24 permute 3) = 24 * 23 * 22, but how many "similar permutations" do we get to eliminate when we try to calculate the combinations of (24 choose 3)? Well, consider one 3-card example: 2 5 K . How many ways can you arrange (permute) these three cards into your three-card hand? Well, from this three-card group, you can place any of the 3 cards in position 1, 2 cards in position 2, and one remaining card in position 3. So there are 3 * 2 * 1 = 6 "similar permutations". So (24 choose 3) must be equal to [ (24 permute 3) / 6 ] = 2024.

If you want to further generalize, it's convenient to use the shorthand notation of factorials. For big N, N! means (N) * (N-1) * (N-2) * ... (3) * (2) * (1). 1! is just one, and 0! is also one. We saw earlier that (24 permute 3) = 24 * 23 * 22, which looks something like the first three terms of 24 factorial ( 24! ). In fact, the generalization is that [N permute R] is [ N! / (N-R)! ]. If you write out some examples, you'll see that all the denominator terms
from N-R down to one cancel the same term in the numerator.
Trying (N,R) = (9,2) for example, shows that (9 permute 2) is (9! / 7!) which is just 9*8.

Notice when we calculated (24 choose 3) we divided the
permutations by 3! = 6. The general rule is that (N choose R) is [ (N permute R) / R! ], and given what we know about permutations now, it's also [ N! / ( (N-R)! R! ) ].

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