SOLUTION: 1.how many ways can 5 people occupy 8 chair if no two people are to occupy the same chair? i do not understand why this question have to use permutation instead of combinations.

Question 412390: 1.how many ways can 5 people occupy 8 chair if no two people are to occupy the same chair?
i do not understand why this question have to use permutation instead of combinations. >.<"
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
permutations assume that order is important.

combinations assume that order is not important.

you have a total of 8 chairs.

5 people can occupy these chairs in P(8,5) = 8! / 3! = 6720 ways.

That's a lot of ways.

To make it simpler to see, let's assume there are 3 chairs and 2 people.

In how many ways can those 2 people occupy 3 chairs.

The answer should be 3! / 1! = 6 ways.

Let's label the chairs a, b, and c.

Now we draw the possible number of sets of 2 from 3 chairs.

They would be:

ab
ac
ba
bc
ca
cb

Notice that the 6 possible ways include sets that involve the same people, only in a different order.

ab can be paired with ba
ac can be paired with ca
bc can be paired with cb

If we were dealing with combinations, then the answer would be 3 possible ways because we only have 3 sets where the people involved in each set are different from the other sets.

you would have:

ab
ac
bc

Order, in this case, would not be important, so you would have less possible sets.

None of these sets have the same 2 people in them.

Formula for permutations is n! / (n-x)!

Formula for combinations is n! / ((n-x)! * x!)

With permutations the answer is 3!/1! = 3*2*1/1 = 6

With combinations the answer is 3!/(1!2!) = 3*2*1/1*2 = 3

The answer is because, when you arrange people in chairs, order is important.

If I did my math correctly, your answer should be 8! / 3! = 6720 possible ways.

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