There is a multi-apartment building with $3$ stories and $4$ apartments at each story. In each apartment lives one person. Three random inhabitants of this building are standing outside the building. What the probability that each of them live on a separate floor (event B).

I want to solve this problem using combinatorics approach. The answer that I have in my book is:

$$P(B) = \frac{|B|}{|\Omega|} = \frac{12 \cdot 8 \cdot 4}{12 \cdot 11 \cdot 10}$$

To best of my understanding the logic goes as follows:

(a) total number of possibilities: ordered sample ($3$ out of $12$)

(b) needed possibilities: first we take a person from any apartment ($12$ possibilities), then a person from two remaining floors ($8$ possibilities) and finally a person from one remaining floor ($4$ possibilities).

My question: When I choose people in this problem I do not think we care about the order. So, I think the sample should be unordered. Probably mathematically it does not matter because the “order” factor is in both numerator and denominator is the same. But if indeed I solve the problem as the unordered sample, how I calculate the possibilities in numerator ($|B|$)?

Many thanks.

  • $\begingroup$ What do you mean by an unordered sample? It is unordered already. You have three random people. When you count, the order doesn't matter . $\endgroup$ – herb steinberg Aug 2 '18 at 21:19
  • $\begingroup$ This is exactly what I mean. When three people stand outside the building, it does not matter in which order they stand. $\endgroup$ – John Aug 2 '18 at 21:22

Your understanding of the solution is correct.

To do it without taking the order of selection into account, observe that there are $\binom{12}{3}$ ways to select three of the twelve apartments. The favorable cases are those in which one of the four apartments on each floor is occupied by the inhabitants standing outside the building. Hence, the probability that each of the three inhabitants of the the three-story building lives on a different floor is $$\dfrac{\dbinom{4}{1}\dbinom{4}{1}\dbinom{4}{1}}{\dbinom{12}{3}}$$

You should verify that this gives the same probability as the solution stated in the book.

  • 1
    $\begingroup$ As simple as thank. Thanks a lot. $\endgroup$ – John Aug 4 '18 at 8:28

You can indeed solve this problem by counting (unordered) combinations of people. In this case the numerator would be the number of ways to choose $3$ people out of the $12$ such that exactly one person lives on each floor. Since $4$ people live on each floor, that would be:


  • $\begingroup$ What would the denominator be? $\endgroup$ – fleablood Aug 2 '18 at 22:23
  • $\begingroup$ The denominator would be the number of ways of choosing 3 from 12, which is 220. $\endgroup$ – alcana Aug 2 '18 at 22:30

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