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?

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