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:
2x + 4y
$40
Step-by-step explanation:
Equation for x small balloons and y large balloons
2x + 4y
Total cost of 10 small balloons and 5 large balloons
x = 10, y = 5
2(10) + 4(5)
= 20 + 20
= $40
M=(6-1)/(8-5)=5/3
y=5x/3+b, using (5,1)
1=25/3+b
b=-22/3
y=(5x-22)/3
...
m=(8-3)/(-1--4)=5/3...
y=5x/3+b, using (-1,8)
8=-5/3+b
b=29/3
y=(5x+29)/3
Since they have the same slope and difference y-intercepts they are parallel lines and will never intersect.
