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How do I know when to use nCr button and when to use nPr button on my calculator?

nCr= combinations I believe NPR= permutations

Is there a general rule I can use to figure out which one to use and when according to the given question?

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In permutations, order counts.
So, if I wanted to know the number of ways to arrange a set of four unique books out of a total supply of 20 books, I'd use $20$ nPr $4$. (Order counts in arranging books.)

In combinations, order doesn't count.
So, if I wanted to know the number of ways to make a team of $4$ people out of a total group of $20$ people, I'd use $20$ nCr $4$. This is because order doesn't matter in choosing a team.

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  • $\begingroup$ @ anorton how do you know that order does or doesn't matter? Ie: how many ways can I get a license plate with different plate combinations. Here the order will matter right? $\endgroup$ – MethodManX Jun 25 '13 at 23:24
  • $\begingroup$ Yes. Because the license tag ABC123 is much different to the police officer pulling me over than the tag BCA321. The best way to tell if order matters is to imagine the situation in your head the best way you can. $\endgroup$ – apnorton Jun 25 '13 at 23:30
  • $\begingroup$ Oops... forgot to @-notify you, @MethodManX (I don't know if you're notified of comments on answers to your questions) $\endgroup$ – apnorton Jun 25 '13 at 23:48
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If you want to calculate ${}_nC_r=\frac {n!}{r!(n-r)!}$, use the nCr button.

If you want to calculate ${}_nP_r=\frac {n!}{(n-r)!}$, use the nPr button.

It is my strong belief that you should not be using either of these buttons until you have a clear understanding of why one of the above formulae is in fact something you want to calculate.

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In brief, combinations should be used in situations where the order doesn't matter and permutations should be used where the order does.

Say I am analysing a 49 ball lottery in which 6 balls are drawn. If I want to work out the odds of jackpot, then I need the number of different possible sets of 6 balls that can be drawn, which is 49C6. So the odds are 1/(49C6)

In contrast say I am a stage magician 'predicting' the lottery results ahead of selection, then I need to get the balls in the right order and the number of different ordered selections is 49P6, so the odds of getting this correct are 1/(49P6).

Incidentally this last number is about 0.00000001%, so if you do see a stage magician do this then you'd be justified in thinking there was something fishy about it.

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  • $\begingroup$ @ TomHarris : Shouln't the lottery numbers be nPr .. because if the numbers are in the right order. The prize is bigger.... $\endgroup$ – MethodManX Jun 26 '13 at 1:51
  • $\begingroup$ Depends on the specific lottery rules. $\endgroup$ – DJohnM Jun 26 '13 at 2:39
  • $\begingroup$ @MethodManX I was going by the UK lottery rules, I didn't know the rules were different in your country. $\endgroup$ – tharris Jun 26 '13 at 12:42

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