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October 18th, 2018, 01:44 PM   #11
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YEA!! Thanks mathman....
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October 18th, 2018, 03:37 PM   #12
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Originally Posted by mathman View Post
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 $\frac{9!}{(3!)^3}$.
OK now I'm confused. If you say the cars are a,b,c and then b,c,a then aren't you implying that the cars are indistinguishable (the order doesn't matter)?

If the cars are distinguishable, shouldn't that make a difference? If your first selection is persons 1,2,3 then doesn't it matter if they go in the red car as opposed to the green car?
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October 19th, 2018, 08:58 AM   #13
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OK now I'm confused. If you say the cars are a,b,c and then b,c,a then aren't you implying that the cars are indistinguishable (the order doesn't matter)?

If the cars are distinguishable, shouldn't that make a difference? If your first selection is persons 1,2,3 then doesn't it matter if they go in the red car as opposed to the green car?
The cars are distinguishable - that's why I labelled them a,b,c in the example. The point is that no matter how you order the cars, any one order will lead to all possible arrangements of people in cars. Rearranging the order will get the same set of people in cars possibilities.
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