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
6.5*10^-5 is the answer if tou have to move the decimal to the right is is negative and then count how many spaces
Answer:
5x + 3y - 4
This expression has three terms (5x, 3y, and -4), 2 different variables (x and y), and a constant (-4)
Step-by-step explanation:
#teamtrees #WAP (Water And Plant)
It has been computed that the probability of choosing two chocolate candies is 2/25.
<h3>How to calculate probability</h3>
The probability of choosing the first chocolate candy will be 2/5.
The probability of choosing the second chocolate candy will be 1/5.
Therefore, the probability of choosing two chocolate candies will be:
= 2/5 × 1/5
= 2/25
Learn more about probability on:
brainly.com/question/25870256
Answer:
1. Formula: b * h ÷ 2
12 * 3 = 36 ÷ 2 = 18 in^2
2. Formula: b * h ÷ 2
5 * 8 = 40 ÷ 2 = 20 ft^2
I could only help on two. Hope it helps though! =)
