Question 1084476: Suppose you have 10 chairs that you want to put around a circular table.
a) If all the chairs are different colours, in how many ways can they be arranged around the table?
This I get. Its just 1x9x8x7x6x5x4x3x2x1 which would be 362880, right?
b) If all the chairs are the same color but are numbered 1 through 10, in how many ways can they be arranged around the table if the chairs that have even numbers must be kept together.
I don't get this at all.
c) If all the chairs are the same colour but are numbered 1 through 10, in how many ways can they be arranged around the table if the chairs numbered 1 and 2 must not be placed next to each other.
I don't get this at all as well.
d) Suppose 4 chairs are red, 2 chairs are blue, 3 chairs are yellow and 1 chair is green. The chairs are identical expect for colour. In how many ways can they be arranged around the table?
Again, don't get this one bit.
Thank you!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! a) (n! / n) = (n-1)! = 362880 ways
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b) even numbers are kept together (there are 5 even numbers)
the 5 even numbers form a group, so the circular permutation rule tells us that there are 5 (the 5 odd numbers) + 1 group objects and they can be arranged (6-1)! = 120 ways
but there are 5! ways to arrange the even numbers so we use the multiplication rule, therefore
there are 120 * 120 = 14400 ways
:
c) there are 8 numbers + 1 group, using circular permutation rule there are (9-1)! = 40320 ways
but the chairs 1 and 2 are not allowed to be next to each other, so we subtract, therefore,
there are 9! - (2 * 8!) = 362880 - 80640 = 282240 ways
:
d) what are the restrictions on how these chairs can be placed?
student answers that there are no restrictions, therefore
9! / (4! * 2! * 3! * 1!) = 362880 / 288 = 1260 ways
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