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Permutations and Combinations


Eyas

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There are 10 seats in a row in a waiting room. There are six people in the room.

(a) In how many different ways can they be seated?

(b) In the group of six people, there are three sisters who must sit next to each other.

In how many different ways can the group be seated? [6]

Part A is easy and is solved with 10P6. But part B is the one I'm having a problem with.. our teacher never discussed permutations in class.

For part B, the markscheme says this:

Total number of ways = 8 x 3! x 7 x 6 x 5

= 10 080

If someone can either explain the total number of ways when some elements are restricted/repeated that'd be great, or if someone can point me out to a good study guide/site that discusses this.

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For Part A, do you mean 10C6?

For Part B, pretend there are 10 seats like so: _ _ _ _ _ _ _ _ _ _

There are 8 ways for the 3 sisters (out of the 6 people) to sit together. It doesn't matter what order the sisters sit in so you use C and not P.

S S S _ _ _ _ _ _ _

_ S S S _ _ _ _ _ _

_ _ S S S _ _ _ _ _ etc

so 8 x 6C3 = 8 x 3!

There are now 7 seats left which the 3 people can sit on in any order (order does matter).

7P3 = 7! / 4! = 7 x 6 x 5

Multiplying the two together gives 8 x 3! x 7 x 6 x 5 = 10 080

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Depends which way you look at it.

10P6 can both stand for:

-the number of ways in which 6 people can occupy 10 different seats

-the number of ways in which 6 seats can be assigned to 10 different people

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Depends which way you look at it.

10P6 can both stand for:

-the number of ways in which 6 people can occupy 10 different seats

-the number of ways in which 6 seats can be assigned to 10 different people

Oh, ok. Thank you for informing me of this.

Edited by Riptide
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