Write the number of permutations in factorial form. Then simplify. J K L M N O P

Mathematics

1 answer:

guajiro [1.7K]3 years ago

7 0

Answer:

The number of permutations of the group of letters given (J, K, L, M, N, O and P) is 7! = 7 . 6 . 5 . 4 . 3 . 2 . 1 = 5,040.

Step-by-step explanation:

The number of permutations without repetition is the number of possible sequences that can occur if we take the letters one by one and is calculated with the factorial of the total number of letters in the group.

Before starting we have 7 possibilities, one for each letter. Once the first letter is chosen randomly, we will have 6 possibilities, after taking the third letter we will have 5 possibilities, and so on.

In total the total number of possibilities to obtain the 7 letters in a certain order are 7 by 6 by 5 by 4 by 3 by 2 by 1, sayed in other words factorial of 7 (7!). The result of the multiplication gives 5,040.

In short we can get the seven letters in 5,040 different sequences.

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The solution is where the two graphs intersect, which is
$(-3,1).$

EXPLANATION

The given system of equations are
$y = \frac{1}{3} x + 2$
and

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We need to graph the two equations.

Let us graph

$y = \frac{1}{3} x + 2$
first.

We need at least two points.

You can choose any appropriate value for x and solve for y. Choosing zero makes our working easier. So let us plot the intercepts.

When
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So this gives us the ordered pair,

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$y = 0$
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We plot these two points and draw a straight line through them to obtain the blue graph in the attachment.

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$y = - x - 2$

We again find the intercepts and plot them.

When
$x = 0$

$y = - 0 - 2$

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This gives the ordered pair

$(0,-2)$

Also, when
$y = 0$

then we have,

$0 = - x - 2$

$2 = - x$

$x = - 2$

Then we again have the ordered pair,

$(-2,0)$

We plot these two points on the same graph sheet to obtain the red graph above.

The intersection of the two lines is
$(-3,1)$

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