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zavuch27 [327]
2 years ago
8

7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many

Mathematics
1 answer:
BabaBlast [244]2 years ago
5 0

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>
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masya89 [10]
<h3>Answer: Choice B</h3><h3>y = x^2 + 7x + 1</h3>

======================================

Proof:

A quick way to confirm that choice B is the only answer is to eliminate the other non-answers.

If you plugged x = 1 into the equation for choice A, you would get

y = -x^2 + 7x + 1

y = -1^2 + 7(1) + 1

y = -1 + 7 + 1

y = 7

We get a result of 7, but we want 9 to be the actual output. So choice A is out.

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Repeat for choice C. Plug in x = 1

y = x^2 - 7x + 1

y = 1^2 - 7(1) + 1

y = 1 - 7 + 1

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We can eliminate choice C (since again we want a result of y = 9)

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Finally let's check choice D

y = x^2 - 7x - 1

y = 1^2 - 7(1) - 1

y = 1 - 7 - 1

y= -7

so choice D is off the list as well

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The only thing left is choice B, so it must be the answer. It turns out that plugging x = 1 into this equation leads to y = 9 as shown below

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