At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Answer:
1,968
Step-by-step explanation:
Let x₁ and x₂, y₁ and y₂, and z₁ and z₂ represent the 3 pairs of siblings, and let;
Set X represent the set where the siblings x₁ and x₂ sit together
Set Y represent the set where the siblings y₁ and y₂ sit together
Set Z represent the set where the siblings z₁ and z₂ sit together
We have;
Where the three siblings don't sit together given as [tex]X^c[/tex]∩[tex]Y^c[/tex]∩[tex]Z^c[/tex]
By set theory, we have;
[tex]\left | X^c \cap Y^c \cap Z^c \right |[/tex] = [tex]\left | X^c \cup Y^c \cup Z^c \right |[/tex] = [tex]\left | U \right | - \left | X \cup Y \cup Z \right |[/tex]
[tex]\left | U \right | - \left | X \cup Y \cup Z \right |[/tex] = [tex]\left | U \right | - \left (\left | X \right | + \left | Y\right | + \left | Z\right | - \left | X \cap Y\right | - \left | X \cap Z\right | - \left | Y\cap Z\right | + \left | X \cap Y \cap Z\right | \right)[/tex]
Therefore;
[tex]\left | X^c \cap Y^c \cap Z^c \right | = \left | U \right | - \left (\left | X \right | + \left | Y\right | + \left | Z\right | - \left | X \cap Y\right | - \left | X \cap Z\right | - \left | Y\cap Z\right | + \left | X \cap Y \cap Z\right | \right)[/tex]
Where;
[tex]\left | U\right |[/tex] = The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!
[tex]\left | X\right |[/tex] = The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!
[tex]\left | Y\right |[/tex] = The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!
[tex]\left | Z\right |[/tex] = The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!
[tex]\left | X \cap Y\right |[/tex] = The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!
[tex]\left | X \cap Z\right |[/tex] = The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!
[tex]\left | Y \cap Z\right |[/tex] = The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!
[tex]\left | X \cap Y \cap Z\right |[/tex] = The number of ways x₁ and x₂, y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 2 × 4!
Therefore, we get;
[tex]\left | X^c \cap Y^c \cap Z^c \right |[/tex] = 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)
[tex]\left | X^c \cap Y^c \cap Z^c \right |[/tex] = 5,040 - 3072 = 1,968
The number of ways where the three siblings don't sit together given as [tex]\left | X^c \cap Y^c \cap Z^c \right |[/tex] = 1,968
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
