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:
p=1/8 q=3/4
Step-by-step explanation:
-2p+7q=5
-6p+35q=51/2
cancel out a variable
(-5)(-2p+7q=5)
-6p+35q=51/2
10p-35q=-25
-6p+35q=51/2 =
4p=1/2
4p/4=1/2 /4
p=1/8
substitute p=1/8 into one of the equation to find q
-2p+7q=5
-2(1/8)+7q=5
-1/4+7q=5
7q=5+1/4
7q=21/4
q= 21/4 /7
q=3/4
This is what the graph would look like. Hope this helps:)
Hi!
<u>Positive</u> <u>discriminants</u> will give you <u><em>two</em></u> <u>solutions</u>.
<u>Discriminants</u> <u>equal</u> <u>to</u> <u>zero</u> will give you <u><em>one</em></u> <u>solution</u>.
<u>Negative</u> <u>discriminants</u> will give you <u><em>no</em></u> <u>solutions</u>.
In a graph, the number of solutions is where the graph crosses the x-axis.
In the first graph, we can see it intersects the graph at two points: (2, 0) and (6, 0). Since there are two solutions it is a positive discriminant.
In the second graph, we can see it intersects at one point: the origin, or (0, 0). Since there is one solution it is a discriminant equal to zero.
In the third graph, we can see it doesn't intersect; it is above the x-axis. Since there are no solutions it is a negative discriminant.
<u><em>For similar problems, see:</em></u>
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