combinations and permutations?

Question

how do you know when you are supposed to use combinations and when you are supposed to use permutations...they dont seem so different to me so i keep getting confused on which one should i use when?

im only in middle school learning pre algebra so its not something that should be complicated to understand. pls help =]

how do you know when order matters or doesnt?

Answer 1

Permutations are used when order matters while combinations are used when order does not matter.
If you are arranging people in a row and want to know how many ways to arrange them, use combinations.
If something has to be in a specific position, use a permutation.

If you are arranging the letters A,B,C into groups of 2, where AB and BA are the same group of letters, it's a combination. If AB and BA are different, it's a permutation.

If you are throwing 2 dice and one is a 3 and the other 4, it's a combination. If the dice are different colors and the red 3 is different from the blue 3, it's a permutation.

Answer 2

One "uses" combinations when you are counting the number of subsets of some specific size from a set, and you "uses" permutations when you are trying to count the number of ordered subsets of some specific size from a set. Let me give you an example: Suppose that you have 10 children, and you are making them compete for your love. You are going to give them prize4s: First prize is $100, second prize is $50 and third prize is $10, and everyone else gets nothing. Then how many ways can the prizes be distributed? Well you need to choose the 3 kids to get the prizes, but also you need to choose an order for those three kids, since the prizes are different, the order that you give them out matters. So you would use a permutation here, and the answer would be P(10,3) =10!/(10-3)!=10!/7!=10*9*8=720. On the other hand, suppose that instead of prizes, you decided to whip the 3 worst kids to death. So here you again need to choose 3 kids out of the 10 to choose from, but the order doesn't mater, since they are all going to get the same punishment. In this case you would use a combination, and the answer would be C(10,3)=10!/(3!)(10-3)!=10*9*8/3*2*1=120.

Answer 3

Let's assume you have a penny, a nickel, a dime, and a quarter, which you put into a hat.

If we ask the question, "How many different ways are there to choose two coins?", we could mean one of two different things. Either we care about the ORDER in which we select the items, or we don't care.
If we care about order, then pulling two coins like so:
Penny, Nickel
is considered DIFFERENT than pulling two coins like so
Nickel, Penny

But if we don't care about order, and we only care about the total amount of money we get, the order doesn't matter, and
Penny, Nickel
is considered the same outcome as
Nickel, Penny.

When you care about order, you use a Permutation: "4 choose 2". We have 4 unique objects and we're choosing 2 of them, and the order matters. The formula for Permutations is
n! / (n-x)!
= 4! / (4-2)!
= 24 / (2)!
= 24 / 2
= 12
The total number of possible outcomes is 12, and they are:
Penny, Nickel
Penny, Dime
Penny, Quarter
Nickel, Penny
Nickel, Dime
Nickel, Quarter
Dime, Penny
Dime, Nickel
Dime, Quarter
Quarter, Penny
Quarter, Nickel
Quarter, Dime

However, if we don't care about order, we use a Combination: "4 choose 2". In this case, the formula is
n! / (x! * (n-x)!)
= 4! / (2! * (4-2)!)
= 24 / (2 * 2!)
= 24 / 4
= 6
There are only six unique outcomes in this case, and they are:
Penny, Nickel
Penny, Dime
Penny, Quarter
Nickel, Dime
Nickel, Quarter
Dime, Quarter

Answer 4

Some examples:
Drawing a sample from a lot (order not important). Use Combinations
How many different committees of 3 can be found among 10 people --Combinations

How many different ways can 7 persons be seated at a table (order important) -Permutations
How many different license plates plates can be formed from 26 Latin letters and 10 integers? - permutations