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I'm learning about probability on khanacademy. They teach a certain method (they draw out the spaces) to calculate combinations.

Two Examples:

1.

Take the question "What is the probability to get exactly 3 heads in 5 flips with a fair coin?" for example. It's first necessary to calculate the amount of possibilities when there's exactly 3 heads. Then the method they use is to first draw out the amount of spaces, i.e. 5 flips:

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And then you see that for the first heads, there's 5 possible spaces, then for the second heads there's 4, etc. So it ends up being $\frac{5 \cdot 4 \cdot 3}{3!}$.

I really like this approach of reasoning, because it tends to be more about logic, rather than just applying a formula. But there's something that confuses me, because it seems like there's some ambiguity in this approach. I'll explain why after the next example.

2.

Now in the question "There's 9 people that want to sit on 3 chairs. How many combinations are there?" we use a similar approach.

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For the first chair there's 9 possible people that can sit down. For the second theres 8 and for the third theres 7, So it ends up being $\frac{9\cdot8\cdot7}{3!}$

The problem/question:

The thing I don't understand is that in the first example we multiply the available amount of spaces ($5\cdot4\cdot3$), but in the second example we multiply how many objects can go in each space ($9\cdot8\cdot7$).

See how these are two different approaches instead of one? When I encounter new problems/questions I don't know whether to multiply the available amount of spaces or to multiply the possible amount of objects in each space. Why are there two different approaches? And how can I tell which one I should use in any given scenario?

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  • $\begingroup$ Think of the persons as spaces. There are $9$ of them and at the bottom of $3$ of them a chair (object) will be placed. $\endgroup$ – drhab Jul 27 '15 at 20:23
  • $\begingroup$ @drhab That actually makes sense, "dank je" :) One thing though... Like you pointed out, there's two ways you can look at it. Either the people or the chairs are "spaces". How do I know which one I should pick? In other words, how do I know in a given scenario whether I should use 9C3 or 3C9? $\endgroup$ – user1534664 Jul 27 '15 at 20:29
  • $\begingroup$ I suspect 9C3 stands for $\binom93$. What do you mean by 3C9? Not $\binom39$ I think?.. $\endgroup$ – drhab Jul 27 '15 at 20:33
  • $\begingroup$ I am sorry, but it is $22:36$ here (bij jou ook, denk ik :) and my wife wants me to go to bed. Tomorrow I will have a second look at your question. Cheers. $\endgroup$ – drhab Jul 27 '15 at 20:36
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    $\begingroup$ To avoid mixing up there is only one route: read very careful what is really asked in the question that you are trying to solve. Questions in wich $k$ exceeds $n$ so that the answer is $0$ are very very rare, so the largest number will (almost) always indicate the 'places'. If there are $9$ persons and only $3$ chairs then that allready indicates that a selection of persons ($3$ out of $9$) must be made. $\endgroup$ – drhab Jul 29 '15 at 8:26
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So really my question is:

How do you know we have to treat the 9 people as spaces, rather than the k spaces that must be selected? In other words, how do you know that n=9 instead of n=3, and how do you know that k=3 instead of k=9. I get these two mixed up.

Here's the answer of @drhab.

Think of the persons as spaces. There are 9 of them and at the bottom of 3 of them a chair (object) will be placed.

To avoid mixing up there is only one route: read very careful what is really asked in the question that you are trying to solve. Questions in wich k exceeds n so that the answer is 0 are very very rare, so the largest number will (almost) always indicate the 'places'. If there are 9 persons and only 3 chairs then that already indicates that a selection of persons (3 out of 9) must be made.

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