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
Answer:
Here's the simplest example possible: let's say x + y = 3 and x - y = 1. Solve the second equation for x by adding y to both sides: (x - y) + y = 1 + y. So x = 1 + y. Take that value of x, and substitute it into the first equation given above (x + y = 3).
Tan205° = .466 tan55° = 1.428 tan25° = .466 -tan25° = " so just tan25
Your answer from smallest to largest is:
3/10
3/5
13/20
22/25
