# Teaching permutations, How to?

I posed this question to my niece while teaching her permutations:

Given four balls of different colours, and four place holders to put those balls, in how many ways can you arrange these four balls in the four place holders?

She reverted with $(4!)^2$ while I was expecting to hear $4!$. Her reasoning was so:

I can select the first of any of the four balls in four ways. Having picked one, I put this in any of the four place holders in four ways. Now, I select the second ball from the three remaining ones in 3 ways. I can place it in any of the remaining three place holders in 3 ways.

... and so on to produce $4^2 * 3^2 * 2^2 * 1^2$

How do I explain to her that only $4!$ arrangements are possible and this is regardless of which order she picks the balls?

• Make her actually do it. (With fewer objects) Jun 5, 2012 at 16:36
• @PeterTamaroff: In her solution she both picks one of the four balls and places it in any of the four positions giving $4^2$ possibilities for the first ball and so on. Obviously many of the permuattions given in such a way are equal. Jun 5, 2012 at 16:45
• @PeterTamaroff: She chooses the ball, and chooses in which remaining spot she will place the ball. Jun 5, 2012 at 17:02
• Her solution is correct. Clearly she interprets "in how many ways can you arrange" as talking about the physical act of picking up balls from the bowl they are in, and setting them in place. Sounds very reasonable, answer is $(4!)^2$. "How many arrangements" is different, that clearly refers to the finished structures, have $4!$. Jun 5, 2012 at 19:37
• We have learned conventions through which we interpret the wording of such questions. Someone who is unfamiliar with the conventions may read a into a question a meaning other than the "intended" one. Jun 6, 2012 at 15:00

Demonstrate it inductively. Clearly there’s only one possible ordering of one object, $A$; that’s $1$. Add a second; it can go before ($BA$) or behind ($AB$), doubling the number of possiblities to $2\cdot1$. Add a third: it can go into any of three slots, tripling the number of possibilities: $3\cdot2\cdot1$. At this stage it’s a good idea still to illustrate everything:

$$\begin{array}{c} &&&\square&B&\square&A&\square\\ &&&\color{blue}{C}&&\color{red}{C}&&\color{green}{C}\\ &&\swarrow&&&\downarrow&&&\searrow\\ \color{blue}{C}&\color{blue}{B}&\color{blue}{A}&&\color{red}{B}&\color{red}{C}&\color{red}{A}&&\color{green}{B}&\color{green}{A}&\color{green}{C}\\ \hline \\ &&&\square&A&\square&B&\square\\ &&&\color{blue}{C}&&\color{red}{C}&&\color{green}{C}\\ &&\swarrow&&&\downarrow&&&\searrow\\ \color{blue}{C}&\color{blue}{A}&\color{blue}{B}&&\color{red}{A}&\color{red}{C}&\color{red}{B}&&\color{green}{A}&\color{green}{B}&\color{green}{C} \end{array}$$

Then each of the $3\cdot2\cdot1$ strings of $3$ objects has $4$ slots in which to insert a new one, for a total of $4\cdot3\cdot2\cdot1$ strings, and so on:

$$\square\quad X\quad\square\quad Y\quad\square\quad Z\quad\square$$

• Sincere thanks for the illustrated answer. Jun 6, 2012 at 14:10

Point out that, by this reasoning, there are $4$ ways to arrange two objects in order.

As for explaining, approach it this way. Rather than having holders specified for the balls (the main source of confusion, here) just think of it as drawing balls at random out of a hat, say. In how many distinct orders can the balls can be drawn, assuming one is drawn at a time? Alternately, if you want to keep the previous approach, point out that under her method, there are in fact $4!$ ways (chronological orders of placement) to obtain any given physical arrangement of the balls! Thus, to enumerate the distinguishable physical arrangements of the balls, we find $$\frac{(4!)^2}{4!}=4!,$$ as desired.

• +1 the issue in her logic is that she includes both "a before b" and "b after a" as distinctly different possibilities when they are, in fact, equivalent. Jun 5, 2012 at 17:07
• Excellent point, Matthew. Jun 5, 2012 at 17:53