Let us assume that we have 30 balls( 7 green, 10 black, 13 white).

I was trying to explain to someone how we count the number of possibilities of getting 3 greens, 3 black, and 3 white balls in a situation where the order is not important, and no replacement is allowed.

I did the following drawing.

And then, this person asked me: "Why do we divide each group of $3$ by $3!$, and not the whole group of $9$ by $9!$ ?" Followed by: "And aren't we assuming in the drawing that we have (3g,3b,3w) when we would also want to count points for e.g. (b,w,w,g,b,g,w,b)?"

I wasn't exactly sure how to answer. Maybe this question is better suited for Math.SE... I don't know.

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    $\begingroup$ Start with order being relevant. If the person knows in this case why it's ABC, where A = 7*6*5 and B = 10*9*8 and C = 13*12*11, then it's a matter of dividing by 1*2*3 = 3! to account for the fact that the 3! many ways of permuting the green's, and similarly for the black's and white's. However, for someone just learning this, the example should be something that all the possibilities can be listed in at most two or three minutes. Also, better to start with one group (i.e. only green's), then move on to two groups, before considering the case of more than two groups. (continued) $\endgroup$ Sep 24 '20 at 20:51
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    $\begingroup$ I recommend both of you read over the first three chapters of Mathematics of Choice by Ivan Niven. This book has some excellent discussions of how to approach these types of counting arguments, and often more than one approach is given. $\endgroup$ Sep 24 '20 at 20:53
  • $\begingroup$ I was rushed in my earlier comments, so I didn't have time to give an example. Here's a possible way of beginning the discussion. Let $a,$ $b,$ $c,$ $d$ be $4$ distinct objects. Have the student write down all $24$ ordered $4$-tuples of these objects in a systematic way (for simplicity omit the parentheses). For instance: $abcd,$ $abdc,$ $acbd,$ $acdb,$ $adbc,$ $adcb,\,\ldots$ Here the system is $ab$ followed by the $2$ ways of permuting the remaining objects, then $ac$ followed by $\ldots,$ then $ad$ followed by $\ldots,$ (continued) $\endgroup$ Sep 25 '20 at 9:19
  • $\begingroup$ and at this point notice what we have is $a$ followed by the $6$ ways of permuting the remaining objects. When finished, there will be $24$ $4$-tuples. Now consider what effect the assumption $b=c$ will have. In each of the $4$-tuples have the student cross out each of the $c$'s and write a $b$ just over where the $c$ was. Now some of the $4$-tuples will be the same. How many distinct $4$-tuples do we now have? Notice how the duplicates correspond to "interchanging $b$ and $c$". (continued) $\endgroup$ Sep 25 '20 at 9:19
  • $\begingroup$ Now perhaps ask if this is the same as the number of $3$-tuples of $3$ objects (seems reasonable before checking, since we now have the $3$ objects $a,$ $b=c,$ and $d).$ Why isn't it? Note that for the $4$-tuples, $a$ can come between the two $b$'s, but not when considering $3$-tuples. And continue elementary investigations like this, BEFORE exhibiting elaborate formulas to be memorized. $\endgroup$ Sep 25 '20 at 9:20

It looks to me like your student just has a different, but valid, idea for approaching the problem. If your student made both of the suggested changes, (1) including all ways of ordering the colors, and (2) dividing by $9!$ rather than $(3!)^3$, they would get the same answer as you got.

The main downside to your student's approach is that including all ways of ordering the colors complicates the solution somewhat. Let's review how this would go. In the first step, we want the number of ways of selecting and arranging nine balls such that there are three balls of each color. In order to ensure that the condition on colors holds, assign color labels to nine slots using the labels G, G, G, B, B, B, W, W, W. We need to consider all possible ways of doing this; the multinomial coefficient $\binom{9}{3,3,3}=\frac{9!}{3!\,3!\,3!}$ is the number of ways of doing so. Multiplying this by the product in the boxes in your drawing, which is the number of ways of putting balls of the appropriate colors in the labeled slots, produces the total number of arrangements of balls: $$ \frac{9!}{3!\,3!\,3!}P(7,3)P(10,3)P(13,3). $$ The second step is to divide this by $9!$, which gives your expression.

The advantage of your method is that it eliminates the complication of having to consider different arrangements of the colors, but you need to convince your student that this is a legitimate procedure. Perhaps it would be better to address whatever issue it was that made you feel the need to present the solution in terms of a series of slots in the first place. Why not simply write down the final expression, $\binom{7}{3}\binom{10}{3}\binom{13}{3}$, directly? A student who has truly internalized both combinations and the multiplication principle should be able to see why that final expression is correct. On the other hand, a student who isn't fully comfortable with the multiplication principle is probably going to have trouble seeing why your procedure, which amounts to factoring out and ignoring the color arrangement aspect of the problem, works.

To convince your student that the final combinations formula is valid, you could try using a tree, but you'll have to reduce the numbers in the problem greatly. You could try starting with four green and three black balls, and looking for the number of selections with one green and two black. The tree may help convince your student that the green selection is independent of the black selection. Label the balls $G1$, $G2$, $G3$, $G4$, $B1$, $B2$, $B3$. The branches in the first layer of the tree the branches are labeled $\{G1\}$, $\{G2\}$, $\{G3\}$, $\{G4\}$. Each of these four branches leads to three branches in the second layer of the tree, labeled $\{B2,B3\}$, $\{B1,B3\}$, $\{B1,B2\}$. The twelve outcomes are obtained by taking the union of the sets along the branches, e.g. $\{G2,B2,B3\}$ when you take the second branch in the first layer and the first subbranch in the second layer. The main point you hope your student will see is that the choices for the black balls remain the same regardless of which green ball was chosen and therefore we need simply multiply the number of green selections by the number of black selections.


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