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October 17th, 2018, 10:28 AM   #1
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Cool Permutations problem

This is a problem that has come up in my math test recently and I'm unsure whether the teacher's solution is correct:

There are 12 people and 3 cars. Each car can fit 4 people. If the owner of the car drives their car, in how many ways could the rest 9 people be distributed in the cars? This would mean that there is one set person per car, so there are 3 spots in every car to fill. The order in which they are sat in the car doesn't matter.

The answer I got is: ((9 above 3)*(6 above 3))/3!
The answer my teacher has is: ((9 above 3)*(6 above 3))*3!

Please leave a detailed response as to which answer is correct because, if mine is correct, I will have to dispute it with her so I need all the details I can get.
Thanks in advance!

Last edited by skipjack; October 19th, 2018 at 11:02 AM.
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October 17th, 2018, 02:51 PM   #2
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What does 9 above 3 mean? Is it $\frac{9!}{6!3!}$?
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October 17th, 2018, 03:37 PM   #3
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Yes, by 9 above 3 i mean 9! divided by 6!*3! but I am not familiar with the formatting on this website.
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October 17th, 2018, 04:20 PM   #4
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Quote:
Originally Posted by Memepolice View Post
The answer I got is: ((9 above 3)*(6 above 3))/3!
The answer my teacher has is: ((9 above 3)*(6 above 3))*3!
To make sure: what is actual results of those 2 calculations?

Last edited by Denis; October 17th, 2018 at 05:14 PM.
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October 17th, 2018, 10:20 PM   #5
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My result is 280 ways, and my teachers is 10080.
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October 18th, 2018, 04:55 AM   #6
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I think the answer depends on whether the cars are distinguished. I'll assume that the cars are somehow distinguished - 1st, 2nd, 3rd or red, green, white, etc.

I get $\displaystyle \binom{9}{3} \cdot \binom{6}{3} \cdot 3!=10080$ which agrees with your teacher.

Call the 9 non-drivers A,B,C,D,E,F,G,H,J. The reason for multiplying by 3! at the end is because we want the arrangement

{A,B,C} {D,E,F} {G,H,J}

to be different from

{D,E,F} {A,B,C} {G,H,J}

since the cars are distinguishable and there are 3! ways of ordering the cars.
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October 18th, 2018, 04:59 AM   #7
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Yeah I see, thank you for clearing that up. She isn’t really good at explaining the permutations, and I remembered a similar task we had in ozr workbook but alas, they were not similar enough.
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October 18th, 2018, 06:39 AM   #8
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Quote:
Originally Posted by Memepolice View Post
My result is 280 ways, and my teachers is 10080.
How did you arrive at 280?

9! / 6! * 3! = 3024

9! / (6! * 3!) = 84
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October 18th, 2018, 09:19 AM   #9
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Changing your problem's wording:
3 boxes are labelled 1,2,3.
9 cards are labelled 1 to 9.
3 cards are put in each box; the order within a box does not matter.
In how many ways can this be done?

Answer is 1680:
[Ways][Box1][Box2][Box3]
[0001][..123][..456][..789]
[0002][..123][..457][..689]
[0003][..123][..458][..679]
[0004][..123][..459][..678]
.....
[1677][..789][..345][..126]
[1678][..789][..346][..125]
[1679][..789][..356][..124]
[1680][..789][..456][..123]
So 1680 ways if order does not matter within box,
but the boxes are distinguishable.

1680 / 6 = 280 : your answer
1680 * 6 = 10080 : teacher's answer

So you guys get together and decide how
to fix the terrible wording of the problem!
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October 18th, 2018, 02:22 PM   #10
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I believe both of you are wrong. The correct answer is $\binom{9}{3}\times \binom{6}{3}=\frac{9!}{(3!)^3}$. You teacher's derivation is almost correct, but the final multiplication by$3!$ is wrong. Rearranging the order the cars doesn't add any more possibilities. Example: cars are a,b,c and people 1-9. Let us use cars in order a,b,c initially and b,c,a alternatively. Consider a sort with (1,2,3) in b, (4,5,6) in c, and (7,8,9) in a. However in the initial sort we have the possibility of (7,8,9) in a, (1,2,3) in b, and (4,5,6) in c. This shows that the initial car order covers all possibilities.

An alternative way of getting the result is considering all possible permutations of 9 people $9!$ and place them 3 at a time in the cars. Since the order within a specific car does not matter, you need to divide by $3!$ for each car, ending up with $\frac{9!}{(3!)^3}$.
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