Jason L. answered • 04/18/17
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The key to knowing if you should use the perm. or comb. formula is if the order matters or not.
If order matters, then you would use a permuatation. So for instance, if you picked green marble #1 and green marble #2 specifically, that would not be the same than had you picked green #2 and green #1 in that order.
If order does not matter, then you would use combinations. Using the last example, this would mean that it does not matter which members of a subset are selected regardless of their order of selection. So green #1 and green #2 would be the same thing as green #2 and green #1, and thus should not be counted twice.
So having all of that in mind for your specific questions, there is a key word in the problem that would tip you off as to whether order matters or not: the marbles are all identical. So picking any 2 greens would be the same as any other 2 greens because they're all the same. That means we're using combinations.
6C4 = 6!/(2! * 4!) = 15
Does this help you solve #2?
Caitlyn L.
Hi, i just checked the answer and your answer is unfortunately wrong :(
I think you forgot to include the fact that only 4 out of the 6 marbles were chosen.
So I think this is a question with different cases suitable for the answer, and the total number of arrangements is the different cases add together.
The correct answer is:
Case 1: 2 red, 2 green chosen = 4!/(2! * 2!)
Case 2: 1 red, 3 green chosen = 4!/(3! * 1!)
Case 3: 0 red, 4 green chosen = 4!/4! = 1
Total no. of arrangements = 1 + 4!/(3! * 1!) + 4!/(2! * 2!) = 11 different arrangements.
The answer said 11, so I did my best to work it out. Please tell me if I'm wrong
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04/28/17
Jason L.
I'm going to be totally honest and say I have no idea where 11 comes from. I'm not saying it isn't right somehow, but the math you wrote out wouldn't make any sense, combinatorically. For instance
"Case 2: 1 red, 3 green chosen = 4!/(3! * 1!)"
I'm not sure what this means? The numbers you wrote out equates to 4C3, but that would not have any meaning because then you're not picking from the specific subset of each color of marble (1 from the 2 reds and 3 from 4 greens). You'd only be counting the 3 from the greens in that equation. You have to account for the 2 possible reds.
This is actually what I believe you were trying to do:
2C2 * 4C2 = 1 * 6 = 6
2C1 * 4C3 = 2 * 4 = 8
2C0 * 4C4 = 1 * 1 = 1
8+6+1 = 15 = 6C4
I'm stumped if the answer says 11 because the math leads to 15. It's possible there's some trick to this question that I am missing though. Consider re-posting this (or maybe emailing your professor). I'd be curious to see what it actually is.
However, if the question is actually asking "how many combinations of colors of marbles" then 11 would be correct because order would no longer matter
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04/28/17
Jason L.
(ignore that last line. I thought I figured out why, but then realized that way still lead back to 15. I just evidently forgot to delete it)
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04/28/17
Caitlyn L.
Oh my haha, my brain is actually not functioning. :(
I'll repost and maybe email my professor as well, but yeah, the answer is 11 apparently.
I'm so confused right now and I have an exam on these soon - I'm dead meat :(
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04/28/17
Jason L.
I actually sent this to my old stats professor after I responded an hour ago. He told me he got 15 as well.
I guess my advice to you is "process over results". The book might be wrong or it might be right (we don't know!). The best thing you can do is familiarize yourself with the formulas and when they are used, so that you'll be able to think these questions through on your exam.
Do you at least understand how we got 15 in this example?
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04/28/17
Caitlyn L.
04/28/17