In celebration of Chinese new year, you are arranging name cards for 16 people to sit around this table. Since the cards are placed on the rotating lazy Susan, the specific chair of a guest does not matter, only who is to their left and right. That is, two namecard arrangements are considered different iff there is a guest whose person on their right or left changes. Scenario 1: Count the number of possible namecard arrangements, if there are no restrictions on who sits next to whom. Scenario 1g: Once the names are placed, you hide 4 golden tickets, one under each of four different namecards. How many ways could you choose to do so? Scenario 2: You aim to encourage bipartisan dialog by inviting 8 guests each from two political parties. Count the number of arrangements in which no members of the same party are in adjacent seats. Scenario 2g: Now when you hide four golden tickets under the placed namecards, you want to ensure that each party has two winners. In how many ways can you choose to do so? Scenario 3: You have invited 8 couples, and couples must not be separated, but must sit together. In how many ways can you choose to arrange the namecards? Scenario 3g: You hide four golden tickets under placed namecards again, but ensure that no couple wins two tickets. In how many ways can you choose to do so? A2RF
In celebration of Chinese new year, you are arranging name cards for 16 people to sit around this table. Since the cards are placed on the rotating lazy Susan, the specific chair of a guest does not matter, only who is to their left and right. That is, two namecard arrangements are considered different iff there is a guest whose person on their right or left changes. Scenario 1: Count the number of possible namecard arrangements, if there are no restrictions on who sits next to whom. Scenario 1g: Once the names are placed, you hide 4 golden tickets, one under each of four different namecards. How many ways could you choose to do so? Scenario 2: You aim to encourage bipartisan dialog by inviting 8 guests each from two political parties. Count the number of arrangements in which no members of the same party are in adjacent seats. Scenario 2g: Now when you hide four golden tickets under the placed namecards, you want to ensure that each party has two winners. In how many ways can you choose to do so? Scenario 3: You have invited 8 couples, and couples must not be separated, but must sit together. In how many ways can you choose to arrange the namecards? Scenario 3g: You hide four golden tickets under placed namecards again, but ensure that no couple wins two tickets. In how many ways can you choose to do so? A2RF
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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In celebration of Chinese new year, you are arranging name cards for 16 people to sit around this table. Since
the cards are placed on the rotating lazy Susan, the specific chair of a guest does not matter, only who is to
their left and right. That is, two namecard arrangements are considered different iff there is a guest whose
person on their right or left changes.
Scenario 1: Count the number of possible namecard arrangements, if there are no restrictions on who
sits next to whom.
Scenario 1g: Once the names are placed, you hide 4 golden tickets, one under each of four different
namecards. How many ways could you choose to do so?
Scenario 2: You aim to encourage bipartisan dialog by inviting 8 guests each from two political parties.
Count the number of arrangements in which no members of the same party are in adjacent seats.
Scenario 2g: Now when you hide four golden tickets under the placed namecards, you want to ensure
that each party has two winners. In how many ways can you choose to do so?
Scenario 3: You have invited 8 couples, and couples must not be separated, but must sit together. In
how many ways can you choose to arrange the namecards?
Scenario 3g: You hide four golden tickets under placed namecards again, but ensure that no couple wins
two tickets. In how many ways can you choose to do so?
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Step 1
Given
Scenario 1
The total number of people
The total number of different arrangements of n people in a circle can be done in (n-1)! ways
The total number of different arrangements of 16 name cards on a circular table assuming no restrictions on who sit next to whom
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