Often in basic mathematics, we can visualize things very easily, which I believe helps understanding (instead of just working out a number theoretical proof). For example: $$(n+1)^2 - n^2 = (n+1) +n$$

can be visualized by squares. Remove a square with sided $n$ from a square with sides $n+1$ leaves the top row ($n+1$) and the right row without the top ($n$) (done here with diamonds and bullets for $n = 4$).

$$ \diamond \diamond \diamond \diamond \diamond \\ \bullet \bullet \bullet \bullet \diamond \\ \bullet \bullet \bullet \bullet \diamond \\ \bullet \bullet \bullet \bullet \diamond \\ \bullet \bullet \bullet \bullet \diamond $$

Another example is proving that $$\sum_{i = 1}^n 2\cdot i = n^2 + n$$ which can be done in the following way (for $n = 4$):

$$ \diamond \diamond \diamond \diamond \\ \diamond \diamond \diamond \bullet \\ \diamond \diamond \bullet \bullet \\ \diamond \bullet \bullet \bullet \\ \bullet \bullet \bullet \bullet $$

Here, we see two triangles, the one with diamonds with row lengths from $1$ to $n$ and the one with bullets going from $1$ to $n$, which represents the sum. We also see a $(n+1) \times n$ rectangle, which represents the right hand side. This proves the theorem.

I was working through same basic number theory proofs and induction proof because I like to visualize these. It is easy enough to visuale $n^a$ as an $a$-dimensional cube with sides $n$. The problem is that I have often difficulty to visualize the factorial: $n!$

Does anybody know of a nice way to visualise the factorial?

The best I could come up with is the following:

See $2!$ as just two dots $\bullet \bullet$.

See $3!$ as a triangle with the sides made with $2!$, e.g. $$ \cdot \\ \bullet \quad \bullet \\ \bullet \quad \quad \bullet \\ \cdot \space \space \bullet \bullet \space \space \cdot $$

Now see $n!$ as an $n$-gon with the sides made of the $(n-1)$-gon. (So $4!$ would be a square with a $3!$-triangle on its sides.)

This visualization is not very easy to work with when you want to visualize proofs. Are there better ways to visualize $n!$?

EDIT: I should emphasis that I would like to visualize $n!$ using dots or lines or so, not so much with concepts ( it is definitely easier to understand the factorial using permutations, just as it is easier to prove some statements using algebra, however the point is that I am trying to prove these things using these very concrete and real visualizations.)

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    Maybe the number of ways to order $n$ books? – wythagoras Aug 03 '15 at 15:47
  • Visualizing $n!$ can't be easy as it grows very fast. –  Aug 03 '15 at 15:49
  • If I recall correctly, it's not just an $(n-1)$-gon, it's actually the next-higher dimension's "three-sided" shape, i.e., the progression is "point, line, triangle, tetrahedron, $4$th-dimension tetrahedron, ..." In particular, it directly follows the progression of Pascal's Triangle for the $n$ number of terms in $(a_1+ a_2+a_3+\dots+a_n)^k$. – abiessu Aug 03 '15 at 15:49
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    Im looking for a more concrete visualization, like with dots. – Krijn Aug 03 '15 at 15:49
  • @abiessu I do not see why that would be the case. The number of dots in this $2$D-representation is $n!$, as we just multiply our $(n-1)!$-construction $n$ times. – Krijn Aug 03 '15 at 16:30
  • Part of the reason that the increasing-dimension visualization appears is because it arises naturally as a consequence of the relationship with Pascal's Triangle, and the fact that higher term counts in the evaluation of $(a_1+ a_2+a_3+\dots+a_n)^k$ take place in higher dimension versions of said triangle. – abiessu Aug 03 '15 at 16:56
  • Ah, I see what you mean now. I would argue that it is a different way of visualizing it than my approach, but I believe it fits my purpose better! – Krijn Aug 03 '15 at 16:59
  • I usually just think of it as the string of symbols "$n!$," and just remember that I can replace $n!$ with $n(n-1)!$ at any time. Either that, or the number of ways of arranging $n$ objects. – Akiva Weinberger Aug 03 '15 at 17:10
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    Perhaps something like [this](http://www.techuser.net/images/5-3-permtree.gif) — a tree with $n$ "children" coming out of the first vertex, $n-1$ "children" coming out of the second, etc. (Or the reverse order.) – Akiva Weinberger Aug 03 '15 at 17:13

4 Answers4


One way is the total number of leaves of a (single) rooted tree in which each leaf is minimally linked to the root by exactly $n-1$ edges, and which has the following property: the root has $2$ children, each child of the root has $3$ children, each child of each child of the root has $4$ children, and so on until the leaves are reached. A natural term for this is factorial tree, but I don't know if this phrase is in general use for this notion.

For example, for $n = 4$: For $n = 4$

Dave L. Renfro
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  • Excellent! Closely related to the comment by@columbus8myhw, although in reverse. – Krijn Aug 03 '15 at 18:05
  • I didn't see @columbus8myhw's comment until just now, or I might not have bothered! Apparently the comment wasn't available just when my lunch hour began, a little over an hour ago when I thought about posting (and eventually decided to post) an answer, and I didn't look back over the comments to see if any new ones had appeared. – Dave L. Renfro Aug 03 '15 at 18:10
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    @DaveL.Renfro: I think you mean "child" rather than "[sibling](http://dictionary.cambridge.org/dictionary/british/sibling)"? – psmears Aug 03 '15 at 20:00
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    @Dave L. Renfro, I have created the factorial tree for $n = 4$, [here](http://i.imgur.com/4g45riR.png). Perhaps it would be useful to display it in your answer. – Krijn Aug 03 '15 at 20:21
  • I don't know how to paste pictures into a stackexchange answer. That was what I meant by "how to do such things in a stackexchange post". – Dave L. Renfro Aug 03 '15 at 20:25
  • @Krijn Now you want to find a natural way to assign to each leaf a permutation. (I think the reverse order works best for this, actually — the first $n$ branches decide the first element in your permutation, the next $n-1$ decide the second element, etc.) – Akiva Weinberger Aug 03 '15 at 23:12

Here's a geometric visualization in higher dimensions. You can take a hyper-cube in dimension $d$ (basically the Cartesian product of $n$ copies of the interval $[0,c]$ for any $c > 0$ that you want), and then you triangulate (i.e. partition) into equal volume simplices (a simplex in $d$ dimensions is a full dimensional convex hulls of $d+1$ points, i.e. higher dimensional analog of triangles for $d = 2$) by first drawing the edge from the origin to the opposite corner of the hypercube (so the opposite corners are vertices included in each simplex), and then move along one edge of the cube incident to the origin to get your next vertex, then move closer to the opposite corner by taking one edge incident to that vertex to get the next vertex, and so on until you reach the opposite corner. You can traverse the dimension-aligned edges in any order you want to get $d+1$ vertices of a distinct simplex, and the interiors of the simplexes are disjoint, and the number of congruent simplexes you get in this partition is equal to the number of ways you can order the dimensions, which is $d!$. Thus, if $c = 1$, then each simplex in this partition has volume $1/d!$ and they are all congruent.

A related construction is to consider the volume of the simplex whose vertices are the origin along with the endpoints of $d$ linearly independent vectors $v_i$ extending from the origin. This solid has the description $\{ \sum_i c_i v_i \, | \, \sum_i c_i \leq 1, c_i \geq 0$ } where the $v_i$ are your vectors. The parallelepiped (analog of hypercube) spanned by these vectors $v_i$ on the other hand has the description $\{ \sum_i c_i v_i \, | \, 0 \leq c_i \leq 1 \}$. It is a geometric fact that the volume of the parallelepiped is $d!$ times the volume of the simplex, and the volume of the parallelepiped is $|\det V|$ where $V$ is the matrix of the vectors that span the parallelepiped.

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  • This is quite a difficult visualization. I do like them (expecially the second one) but was looking for a simpler one, one that would still make sense for a high school student or a first year undergraduate. – Krijn Aug 03 '15 at 16:29
  • @Krijn Fairly early in undergraduate (e.g. in the 1st or 2nd year) one usually has the opportunity to take basic linear algebra and then, at least if they take a geometric approach at times, this should hopefully all be explained. I apologize I couldn't give you an answer closer to what you were looking for. – user2566092 Aug 03 '15 at 16:34
  • Don't apologize! I certainly liked the answer. I however was looking for an answer that would easily explain for example why $n!$ grows faster then $n^a$ or that could prove $\sum_{i=1}^n\frac{i-1}{i!} = \frac{n!-1}{n!}$ using a visual approach. – Krijn Aug 03 '15 at 16:37
  • @Krijn I don't know if I would exactly call it "visual", but just based on the definition of factorial, you can easily show the pattern that gives you your second equation for example, just by adding $1/n!$ to both sides and seeing the pattern of what happens in the computations if you start with $i=n$ and work your way down. That basically has to do with the fact that the factorials can be used as a base system for writing down numbers, which is an interesting fact about them. – user2566092 Aug 03 '15 at 16:46
  • @Krijn Also for your first claim, you can write $(n!)^2$ and imagine visually taking pairwise products of terms with $n!$ written forward as a product in one copy and backwards in the other. Then it's not too hard to see that each pairwise product of terms is greater than or equal to $n$, which shows $n! \geq n^{n/2}$. Visual, just maybe not how you meant visual. – user2566092 Aug 03 '15 at 16:48
  • I am sorry, I might have to use a different word than visual. Indeed it is not that I am trying to proof these equations generally, they were just examples of things that I was hoping I could explain using dots, instead of number theory. – Krijn Aug 03 '15 at 16:57

The way I see $n!$ is a hybrid of avid19's and Dave L. Renfro's visualizations: I imagine $n$ people lining up one by one. I think it really helps to imagine people or animals or fruits or something, rather than boring symbols: that's the way it's done in Burns and Weston's Math For Smarty Pants, and it seems to have made quite an impression on me. My keyboard has no fruits on it, unfortunately, so maybe try to imagine the digits below pinned to some hockey players.

  • The first person doesn't have any choice about where they join the line, since there is no line yet.


  • The second person can join in two places: the front or the back.

    21 12

  • The third person can join in three places: the front, the middle, or the back.

    321 231 213

    312 132 123

  • The fourth person can join in four places.

    4321 3421 3241 3214

    4231 2431 2341 2314

    4213 2413 2143 2134

    4312 3412 3142 3124

    4132 1432 1342 1324

    4123 1423 1243 1234

  • The fifth person can join in five places...
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This might not be what you're looking for, but I visualize a factorial as a process. $5!$ is how many ways you can arrange 5 things. I visualize arranging 5 things. Not a representation as dots but personally it's powerful.

  • You can *visualize *the number of ways one can order $5$ things? I'm impressed. I can hardly do $3$ without trying unnaturally in my head. – MCT Aug 04 '15 at 21:07
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    @soke Not the individual ways of course. But the process of it. "5 things here, then 4, then ..." which is basically the definition of the factorial –  Aug 06 '15 at 23:00