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:
7.18 light-years
Step-by-step explanation:
6.8 x 10^13 = 68,000,000,000,000
9.46 x10 ^12 = 9,460,000,000,000
6.8 x 10^13 * 1 light year / 9.46 x 10^12 = 7.188 light years.
Have a Merry X-MAS Plz MArk Branilest
3/4(x+21) = 10.5
0.75x+15.75 = 10.5
0.75x = -5.25
x = -7
Hope it helps : )
m = -1
-5 = 8m + 12/4
-5 - 12/4 = 8m
4(-5) - (12/4) = 4(8m)
*to remove the denominator, multiply each figure by 4.
-20 -12 = 32m
-32 = 32m
-32/32 = 32m/32
-1 = m
m = -1
To check:
-5 = 8m + 12/4
-5 = 8(-1) + 3
-5 = -8 + 3
-5 = -5
