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Andrassy
Registered User Posts: **777** Member

Could someone explain the difference for me and maybe give an example? nPr is where the probability must be done in a certain way and nCr is any way? i dont really understand this. could someone explain please =)

Post edited by Andrassy on

## Replies to: nPr and nCr

146Junior MembernCr is combination, which is the probability when the order does not matter.

777Member4,706Senior Member146Junior Memberfor example, if you have to find how many diff 'combinations' you can get of 3 out of 12 pizza toppings , then that's just combination because the order in which you put the toppings doesn't matter.

however, if it's a problem like this: in how many ways can be chosen a president, vice president and secretary from a group of 30 students? in this case order does matter since it's not the same choosing a president and choosing a vice president.

i hope you understood, i translated these examples from my math in spanish book. :S

172Junior MembernCr = n! / ((n-r)!*r!)

746Member1,267Senior Member1,012Member408Member#people/things nCr #of spots

but if order matters then

#people/things nPr #of spots

1New MemberOr you can remember that permute means "change the order of".

In addition remember that there will certainly be more ways in which to do things when order matters, hence when you permute, you lose one of the terms in the denominator, thus yielding a larger number of options when you permute.

73Junior Member-Use nPr when order matters. For example, imagine stacking scoops of ice cream in an ice cream CONE. You can choose to put chocolate, vanilla, and strawberry in a certain order since the cone only stacks one way.

However...

-Use nCr when order does NOT matter. For example, imagine placing scoops of ice cream in a BOWL. Here, the order does NOT matter because you can place the ice cream in the bowl without stacking them on top. You can just get different combinations in the bowl without worrying about the order.

2,039Senior MemberI have 7 different colored marbles. I choose 4 from the bag. How many different possibilities are there?

2,039Senior MemberSecond marble can any of the 6 remaining marbles.

Third " " " 5

Fourth " " " 4

Hence: 7*6*5*4 =30*28 =

840But wait, I am overcounting!

Divide by 4!=4*3*2=24

35206Junior Member237Junior Member