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# nPr v nCr

Posted on 2006-04-10
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I understand that you would you use nCr when counting the number of possible ways you can for example be dealt 5 cards out of 52, or 3 out 13. 52 C 5 and 13 C 3. When do you use nPr. What is the difference in terms of when you would use them, and would use ever use nPr with cards for calculating probablitiies?
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Question by:walker6o9

LVL 84

Accepted Solution

You would use nPr if the order of the cards makes a difference.
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LVL 1

Author Comment

Can you please explain further? Why would that mean that there are MORE possibilities.(Since nPr is always a higher number than nCr).
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LVL 84

Expert Comment

nP0 and nP1 are not higher than nC0 and nC1
the number of ways to shuffle a deck is 52P52 not 52C52
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LVL 11

Expert Comment

Hi,
nCr is always <= nPr because it takes into account the ordering of the cards az ozo has written.
So if you have to select 3 cards, let one of the possibilities be 8, 9, 10 for example. Then the ordering of these 3 cards doesn't matter for 52C3, but it does for 52P3. 52P3 will count all different orderings of these 3 cards and they are
8, 9, 10
8, 10, 9
9, 8, 10
9, 10, 8
10, 8, 9
10, 9, 8

Hope that helps!
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LVL 18

Expert Comment

I find it useful to think of nCr = nPr/r!

if you aren't interested in the order, then you divide by the number of orders there are for the r selections (r!)
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LVL 26

Expert Comment

I believe you are correct.  I can't think of a card game where you would use nPr.

You use nCr to look at the number of hands in Contract Bridge or Draw Poker.
These are games where you get all of your cards before you have to take any action.

In games like Texas Hold'em, where the order of the cards matters enormously, you do very different kinds of analysis.

A Draw hand like [2 7 2 2 2] is an almost sure winner.

But you would be crazy to play a Hold'em hand like [2 7][K K 2][2][2] to the end.
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LVL 84

Expert Comment

This is a game in which order makes a difference http://en.wikipedia.org/wiki/FreeCell
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LVL 1

Expert Comment

If you would pick 3 cards out of 52, the number of possibilities would be:

- for the first card: 52
- for the second card: 51
- for the third: 50

so the total number of possible picks would be: 52 * 51 * 50 = 132600 sets of 3 cards

In Math this is called the number of permutations (nPr: 3 out of 52 = 3P52)

BUT, these permutations have a lot of the same set of 3 cards, if you don't mind about the order of your pick.

In these 132600 posibilities (permutations) there are for example 6 different ways to pick 3 Aces. (A1-A2-A3 or A1-A3-A2 or A2-A1-A3 or A2-A3-A1 or A3-A2-A1 or A3-A1-A2). These are alle different permutations of the same combination of those 3 Aces.

So, with permutations the picking order A1 A2 A3 is different then A3 A2 A1. But if you dont mind the order of picking your cards, if the meaning of the order has no relevance and you just want to know about those 3 Aces, you have a lot of permutations that are actualy the same.

As a matter of fact, each set of 3 cards you can pick has 3! = 6 variations (like the 3 Aces)

That is why you have to use Combinations (3C52) when the order of picking has no meaning. That is why you have to devide all the possible sets of 3 cards (permutations) by 3!, to only count the number of unique sets.

REMEMBER
1. Select 7 out of 7 and put them back: 7^7 possibilities
2. Select 7 out of 7 and don't put them back: 7! = 7*6*5*4*3*2*1 possibilities
3. Select 3 out of 7 and don't put them back and mind the order: permutations 3 out of 7 = 7*6*5
4. Select 3 out of 7 and don't put them back and DON'T mind the order: combinations 3 out 7 = 7*6*5 / 3!

Cheers!
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LVL 11

Expert Comment

Hi walker6o9,

Since I see you have not much experience here, you can read this section of the site's help about grades.
Here it is - http://www.experts-exchange.com/help.jsp#hi73

Just for info: If a solution works for you without problems it is normal to give a grade of A according to the site recommendations.
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