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The way I understand it, many uniform probability problems can be solved by looking at either permutations or combinations.

Example problem:

3 people taking the elevator can go out on 9 different floors. Assume equal probability for any person going out on any floor. What is the probability that every person goes out on a different floor?

All possible permutations: Any of the 9 floors can be chosen 3 times = permutations with replacement = $9^3$.

Satisfactory permutations: Permutations without replacement = 9P3.

Probability = $\frac{9P3}{9^3}$ = $\frac{56}{81}$

...which is correct.

Now I thought I could do the same, but with respect to combinations, dividing "combinations without replacement" by "combinations with replacement":

Probability = $\frac{9C3}{11C3}$ = $\frac{28}{55}$

So the answers are not the same. And I don't understand why the latter doesn't work. Knowing the probability and thinking that in the combinations case the denominator should be 11C3, then i calculate that the nominator must be a fraction... which I interpret as there is no way to reach the answer by looking at combinations. Could someone please explain to me whether that is right or wrong, and why?

Sondr
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  • How do you get $11C3$? – Math Lover Aug 28 '20 at 14:01
  • https://math.stackexchange.com/questions/474741/formula-for-combinations-with-replacement – Sondr Aug 28 '20 at 14:34
  • k-permutations and combinations are different, and apply in different situations. As you described, it makes sense to use permutations here. But you can't just say, by analogy, combinations will also apply. The difference is, does the order matter, or not, and here it does. – Todor Markov Aug 28 '20 at 14:57
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    See why is flipping a head then a tail a different outcome than flipping a tail then a head?. The end result here is that the outcomes counted by combinations with repetition (stars and bars) are not equally likely to occur in any reasonable interpretation of this problem and so are not to be used for the calculation of the probability as a ratio of number of good cases to total number of cases. – JMoravitz Aug 28 '20 at 15:33
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    Recall, for event $A$ and sample space $S$ you have $\Pr(A)=\dfrac{|A|}{|S|}$ is only valid when the outcomes in $S$ are equally likely to occur. There are two outcomes to a lottery, you either win or you lose, but you certainly don't win a lottery with probability $\dfrac{1}{2}$. – JMoravitz Aug 28 '20 at 15:34

1 Answers1

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Let's take a simpler version of this question itself. Say there are $2$ persons $A$ and $B$ and only $2$ floors.

The possible permutations {floor 1, floor 2} $= \{AB,\_\},\{A,B\}, \{B,A\}, \{\_, AB\}$.

Each event is equally likely to occur. The arrangements you are interested in are 2nd and 3rd which is them going on different floors. So the probability is $50\%$.

Now let's take the 2nd approach of yours.

The possible combinations of {floor 1, floor 2} $= \{2,0\},\{0,2\}, \{1,1\}$.

You will be interested in the 3rd event and that will give the probability $= \dfrac{1}{3}$ which is wrong because -

$\{2,0\},\{0,2\}, \{1,1\}$ are not equally likely to occur. $\{2,0\}$ and $\{0,2\}$ are each likely to occur $25\%$ of the time. $\{1,1\}$ is likely to occur $50\%$ of the time as both $A$ and $B$ can independently choose their floors.

Math Lover
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