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

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 ∩∩

By set theory, we have;

= =  

=

Therefore;

Where;

= The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!

= The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!

= The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!

= The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!

= The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!

= The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

= The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

= 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;

= 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)

= 5,040 - 3072 = 1,968

The number of ways where the three siblings don't sit together given as  = 1,968

29 mil

step-by-step explanation:

i dont know sorry

step-by-step explanation: