Question

In how many ways can we seat 3 pairs of siblings in a row of 7 chairs, so that nobody sits next to their sibling

Answer 1

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 $X^c$∩$Y^c$∩$Z^c$

By set theory, we have;

$\left | X^c \cap Y^c \cap Z^c \right |$ = $\left | X^c \cup Y^c \cup Z^c \right |$ =  $\left | U \right | - \left | X \cup Y \cup Z \right |$

$\left | U \right | - \left | X \cup Y \cup Z \right |$ = $\left | U \right | - \left (\left | X \right | + \left | Y\right | + \left | Z\right | - \left | X \cap Y\right | - \left | X \cap Z\right | - \left | Y\cap Z\right | + \left | X \cap Y \cap Z\right | \right)$

Therefore;

$\left | X^c \cap Y^c \cap Z^c \right | = \left | U \right | - \left (\left | X \right | + \left | Y\right | + \left | Z\right | - \left | X \cap Y\right | - \left | X \cap Z\right | - \left | Y\cap Z\right | + \left | X \cap Y \cap Z\right | \right)$

Where;

$\left | U\right |$ = The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!

$\left | X\right |$ = The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!

$\left | Y\right |$ = The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!

$\left | Z\right |$ = The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!

$\left | X \cap Y\right |$ = The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!

$\left | X \cap Z\right |$ = The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

$\left | Y \cap Z\right |$ = The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

$\left | X \cap Y \cap Z\right |$ = 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;

$\left | X^c \cap Y^c \cap Z^c \right |$ = 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)

$\left | X^c \cap Y^c \cap Z^c \right |$ = 5,040 - 3072 = 1,968

The number of ways where the three siblings don't sit together given as $\left | X^c \cap Y^c \cap Z^c \right |$  = 1,968