Usually in any problem, whether or not I consider the order of arrangement, my answer is always the same. For example : suppose I have an urn with 2 white and 3 black balls. 2 black balls can be drawn in 6 ways (order taken into account) and any 2 balls can be drawn in 20 ways (order taken into account). So the probability that 2 balls drawn at random are both black is 3/10. Alternatively, 2 black balls can be drawn in 3 ways (order not taken into account) and any 2 balls in 10 ways (order not taken into account). So the probability is 3/10, which is consistent with the previous answer. Today I encountered a problem where the two answers differed.

$The$ $Problem :$ How many ways can 3 balls $(A,B,C)$ be distributed into 3 boxes $(I,II)$ such that box $I$ contains 1 ball?

Order of balls within each box not considered:

I list out all the possibilities - $ABC| $ ; $ |ABC$ ; $AB|C$ ; $BC|A$ ; $CA|B$ ; $A|BC$ ; $B|CA$ ; $C|AB$ ; $ |ABC.$

$2^3$ possibilities. Out of these 3 are favourable to our event. So the probability is 3/8, considering these are all equally likely.

Order considered:

Possibilities - $ABC| $ ; $ACB| $ ; $BCA| $ ; $BAC| $ ; $CAB| $ ; $CBA| $ ; $AB|C$ ; $BA|C$ ; $AC|B$ ; $CA|B$ ; $BC|A$ ; $CB|A$ ; $A|BC$ ; $A|CB$ ; $B|CA$ ; $B|AC$ ; $C|AB$ ; $C|BA$ ; $ |ABC$ ; $| ACB$ : $ |BCA$ ; $ |BAC$ ; $ |CAB$ ; $ |CBA.$

$[3+(2-1)]!$ possibilities. Out of these 6 are favourable to our event. So the probability is 1/4, considering the outcomes are equally likely. Which one is wrong and why? How should I be approaching a problem if in general, the two methods yield different answers?


The first approach gives the correct answer. Specifically, if you want to take order into account, you have to take all the ordering into account, including the order in which the balls are placed. For instance, let's say the boxes are narrow, so that the balls land on top of one another (and that way we can make sense of the concept of an order of the balls in a box), and we pick the balls one by one and throw them in one of the two boxes.

Then $ABC|$ can only happen one way, namely that the balls are thrown in the order $ABC$, and they all go into the left box. However, the possibility $AB|C$ can happen in three ways: Either of the orderings $ABC, ACB, CAB$ give an equal probability of getting $AB|C$. Therefore $AB|C$ is three times more likely than $ABC|$.

If you go over it again with this in mind, you will see that there are not $24$ possibilities, but $8\cdot 6$ possibilities (six ways to order $ABC$, then eight ways to pick boxes for each of them). Of these, $18$ are favourable. The answer is therefore $\frac{18}{48} = \frac38$.

That being said, deciding when to take order into account, and what ordering to take into account and what to ignore is one of the most difficult parts of combinatorics in general. There isn't much I can say that will help you in general, except read the text carefully, be conscious about what ordering assumptions you make, and get practice.

  • $\begingroup$ The possibility $AB|C$ can happen in three ways... I'm sorry I don't quite understand how $\endgroup$ – Hrit Roy Sep 20 '17 at 12:07
  • $\begingroup$ I tried explaining it in the post. First decide what order to throw the balls. For each of the six initial orderings you can choose there are eight possible final outcomes. For instance, if you start with the ordering $ABC$, then you can get $|ABC, A|BC, B|AC, C|AB, AB|C, AC|B, BC|A, ABC$. Each of those are equally likely. However, while $|ABC$ and $ABC|$ can only come about from the initial ordering $ABC$, all the others can come about form two other orderings. For instance, $AB|C$ is also a possible result if you chose the initial ordering $ACB$, or if you chose $CAB$. $\endgroup$ – Arthur Sep 20 '17 at 12:13
  • $\begingroup$ $AB|C$ is therefore three times as likely as $ABC|$ in the end. There are three of the six initial orderings that give you a shot at getting $AB|C$, while there is only one that gives you a shot at getting $ABC|$. $\endgroup$ – Arthur Sep 20 '17 at 12:15
  • $\begingroup$ I understand now. It happens because for $AB|C$ , $C$ can go into box $II$ before $A$, in-between $A$, $B$, or after $B$. $\endgroup$ – Hrit Roy Sep 20 '17 at 14:12
  • 1
    $\begingroup$ @HritRoy Exactly. Which approach is "correct" depends on whether you label the balls before or after you line them up. If you line them up and then put the name $A$ on the first, $B$ on the second and $C$ on the third, then the first approach is what you get. If you put names on the balls first, and then line them up, the second approach is the easiest. $\endgroup$ – Arthur Sep 20 '17 at 14:20

To begin with, your problem statement is really, "When three balls ($A,B,C$) are distributed into two boxes ($I,II$), what is the probability that box $I$ contains exactly one ball?"

If that's not your problem statement, then both your answers are answering the wrong question, since both of them assert a probability.

Since it's a probability question, I would be asking myself how to partition the events so that each event is equally likely. It's not obvious to me whether $AB|C$ should be equally likely, more likely, or less likely than $ABC|$, but it is obvious to me that a reasonable interpretation of the problem is that when it comes time to place ball $A$ (in whatever sequence I place the balls), I should have as much chance to put $A$ in box $I$ as in box $II.$ The chances of the two boxes are similarly equal for $B$ and for $C.$

So I partition all possible events into whether they have ball $A$ in box $I$ or in box $II.$ Those two partitions are assigned equal probability. Then I partition each of those partitions into whether they have ball $B$ in box $I$ or in box $II.$ Now I have four partitions with equal probability. Now I repeat this for ball $C,$ and I have eight equally-likely events.

I observe that I get the same eight events, still equally likely, regardless of the order in which I decide to consider the three balls.

I also observe that three of the eight events have exactly one ball in box $I.$ Therefore the probability is $\frac38.$

How you notate the individual events is a somewhat separate question, but notation can influence how easy it is to recognize whether you listed the events correctly. Rather than either of your notations, I prefer to just make an ordered list of the locations of each of the balls: $A$ first, then $B,$ then $C.$ For example, $A$ in box $II$ and the others in box $I$ is $(II,I,I).$ And just to make things a little neater, I'll rename boxes $I$ and $II$ to $0$ and $1$ so that I can write this arrangement of the balls as $(1,0,0).$ Now I can see that the eight arrangements correspond to the eight integers that can be written with up to three binary digits: $000,$ $001,$ $010,$ $011,$ $100,$ $101,$ $110,$ and $111.$

  • $\begingroup$ Yes. Thinking it like this makes it very clear. $\endgroup$ – Hrit Roy Sep 20 '17 at 14:12

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