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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
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