Question

In a doctor's waiting room, there are 14 seats in a row. Eight people are waiting to be seated. Three of the 8 are in one family, and want to sit together. How many ways can this happen?

Answer 1

There are 14 chairs and  8 people to be seated. But among the 8. three will be seated together:

So 5 people and (3) could be considered as 6 entities:

Since the order matters, we have to use permutation:

¹⁴P₆ = (14!)/(14-6)! = 2,162,160, But the family composed of 3 people can permute among them in 3! ways or 6 ways. So the total number of permutation will be ¹⁴P₆  x 3! 

2,162,160 x 6 = 12,972,960 ways.

Another way to solve this problem is as follow:

5 + (3) people are considered (for the time being) as 6 entities:

The 1st has a choice among 14 ways
The 2nd has a choice among 13 ways
The 3rd has a choice among 12 ways
The 4th has a choice among 11 ways
The 5th has a choice among 10 ways
The 6th has a choice among 9ways

So far there are 14x13x12x11x10x9 = 2,162,160 ways
But the 3 (that formed one group) could seat among themselves in 3!
or 6 ways:

Total number of permutation = 2,162,160 x 6 = 12,972,960