Suppose you're on a 4 × 6 grid, and want to go from the bottom left to the top right. How many different paths can you take? Avoid backtracking -- you can only move right or up.

Spend a few seconds thinking about how you'd figure it out.

Insight: Convert Pictures To Text

When considering the possible paths (tracing them out with your finger), you might whisper "Up, right, up, right...".

Why not write those thoughts down? Using "u" and "r" we can write out a path:

r r r r r r r u u u u

That is, go all the way right (6 r's), then all the way up (4 u's). The path in the diagram would be:

r r r r u u u u r r

Using the text interpretation, the question becomes "How many ways can we re-arrange the letters rrrrrruuuu?"

Ah, the ubiquitous combination/permutation problem -- never thought it'd be useful, eh?

Understanding Combinations And Permutations

There's several ways to see combination and permutation problems. Once the first explanation clicks, we can go back and see it a different way. When trying to build math intuition for a problem, I imagine several mental models circling a core idea. Starting with one insight, I work around to the others.

Approach 1: Start The Same

Instead of having 6 rights at 4 ups, imagine we start with 10 rights (r r r r r r r r r r).

Clearly this won't do: we need to change 4 of those rights into ups. How many ways can we pick 4 rights to change?

Well, we have 10 choices for the first 'right' to convert (see the combinations article). And 9 for the second, 8 for the third, and 7 choices for the final right-to-up conversion. There are 10 * 9 * 8 * 7 = 10!/6! = 5040 possibilities.

But, wait! We need to remove the redundancies: after all, converting moves #1 #2 #3 and #4 (in that order) is the same as converting #4 #3 #2 #1. We have 4! (4 * 3 * 2 * 1 = 24) ways to rearrange the ups we picked, so we finally get:

We're just picking the items to convert (10!/6!) and dividing out the redundancies (4!).

Approach 2: Just Use the Combination Formula

Halfway through that explanation, you might have realized we were recreating the combination formula:

That's the shortcut when you know order doesn't matter. However, sometimes I'm not sure whether I need a permutation or combination from the outset. While saying "Just use C(10,4)" may be accurate, it's not helpful as a teaching tool. Sometimes it helps to re-create the situation on your own.

Approach 3: Start Different

Here's another approach: instead of letting each r and u be interchangeable, label the 'right' moves r1 to r6, and the 'up' moves u1 to u4. How many ways can we re-arrange these 10 items?

This question is easy: 10! = 3,628,800 (wow, big number). We have 10 choices for the 1st move, 9 for the second, and so on, until we have 2 choices for the 9th and only 1 for the last. Cool.

Of course, we know that "r1 r2 u1 u2" is the same path as "r2 r1 u2 u1". We can shuffle the r's and u's in their own subgroups and the path will stay the same.

• How many ways can we shuffle all 10? 10! = 3,628,800
• How many ways can we shuffle 6 r's? 6! = 720
• How many ways can we shuffle 4 u's? 4! = 24

So, we start with the total number of possibilities (10! = 3,628,800) and divide out the cases where we shuffle the r's (6! = 720) and the u's (4! = 24):

Neat! It's cool seeing the same set of multiplications and divisions in different ways, just by regrouping them.

Why is this useful?

One goal is to learn how problems can be transformed. Remember that painting of the old lady & young woman?

Do you see both? Can you switch between them? Isn't that cool?

Part of the fun of the grid-path puzzle is seeing how to look at a problem using a visual or text metaphor. The more math you learn, the more models you have available, and you can turn problems into each other.

This doesn't have to be "practical" -- it's fun to see how listing out paths can be be done simply using letters on paper.

In math lingo, problems which can be converted to each other are "isomorphic". Mathematically, they may be the same -- but from a human perspective, one may be easier than the other (like seeing the old woman or young woman first).

For the grid puzzle, we used each perspective where comfortable:

• Visualizing the grid to understand the general problem and see a single path.
• Write the paths as text to see the general format of all paths & an easy method to enumerate them

And that's the key lesson: It's completely fine to use one model to understand the idea, and another to work out the details. Math becomes difficult when we think there's only one way to approach it.

Variations and Extensions

Now that we've been building our mental models, let's tackle some harder problems.

Imagine your "grid" is actually in 3 dimensions. This is harder to draw, but the text representation keeps on working. Let's say we have a cube (x, y and z dimensions) that is 5 units long on each side. How many paths are there from one corner to its opposite?

Hrm. In this case, I might try the second approach, where we listed out all the possibilities. Assume we label each move differently: we have 5 uniquely-labeled moves of each type (x1-x5, y1-y5, z1-z5). We can arrange these in 15! ways (it's huge: 1.3 trillion). But, we need to remember to divide out the redundancies for each dimension.

There are 5! ways to rearrange the 5 identical motions in each direction, and we divide them out:

Wow, that's huge number of paths on a small cube! Earlier today you'd have trouble with the question -- I know I would have. But starting with the grid example and converting it to text, we've beefed up our model to handle 3 dimensions. Paths in four, five or 10-d should be no problem.

Redefining The Problem

Here's the fun part: instead of changing how we see the solution, why not change the problem? What else could "Find paths on a grid" represent?

• Trap platform: Let's say you're making a set of trapdoors 4 × 6, with only 1 real path through (the others drop you into a volcano). What are the chances someone randomly walks through? With a 4×6 it's 210, as before. With a 12×12 grid it's 24!/12!12! = 2.7 million paths, with only 1 correct one.

• Order of operations: Suppose you have 10 sets of exercises to do: 4 identical leg exercises, and 6 identical arm exercises. How many different routines can you pick? This is the same as navigating the path, except the axis labels are "legs" and "arms" instead of "right" and "up".

• Random walk. Suppose we know an object moves randomly up or right. What's the chance it hits our desired endpoint after 10 steps? Well, there are 2^10 = 1024 ways to move up or right (pick "u" or "r" 10 times), and 210 ways to get to our exact destination. Therefore, you can expect to hit our spot 210 / 1024 = 20.5% of the time!

Here's a calculator to play with a few variations:

Onward and Upward

Puzzles are a fun way to learn new mental models, and deepen your understanding for the ones you're familiar with. While I might "know" combinations and permutations, it's not until I recognize them in the wild do I feel really comfortable. Ideas do no good sitting inside your head like artifacts in a museum -- they need to be taken out and played with. Happy math.

Here’s an easy way to remember: permutation sounds complicated, doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).

Combinations, on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.

Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).

You know, a "combination lock" should really be called a "permutation lock". The order you put the numbers in matters.

A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct.

Permutations: The hairy details

Let’s start with permutations, or all possible ways of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of each item. Let’s say we have 8 people:

1: Alice
2: Bob
3: Charlie
4: David
5: Eve
6: Frank
7: George
8: Horatio

How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)

We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:

• Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Let’s say A wins the Gold.
• Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.
• Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.

We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was $8 * 7 * 6 = 336$.

Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.

We know the factorial is:

Unfortunately, that does too much! We only want $8 * 7 * 6$. How can we “stop” the factorial at 5?

This is where permutations get cool: notice how we want to get rid of $5 * 4 * 3 * 2 * 1$. What’s another name for this? 5 factorial!

So, if we do 8!/5! we get:

And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:

where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have n items total and want to pick k in a certain order, we get:

And this is the fancy permutation formula: You have n items and want to find the number of ways k items can be ordered:

Combinations, Ho!

Combinations are easy going. Order doesn’t matter. You can mix it up and it looks the same. Let’s say I’m a cheapskate and can’t afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.

How many ways can I give 3 tin cans to 8 people?

Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.

This raises an interesting point — we’ve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, let’s just figure out how many ways we can rearrange 3 people.

Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 * 2 * 1$ ways to re-arrange 3 people.

Wait a minute… this is looking a bit like a permutation! You tricked me!

Indeed I did. If you have N people and you want to know how many arrangements there are for all of them, it’s just N factorial or N!

So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we just create all the permutations and divide by all the redundancies. In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.

The general formula is

which means “Find all the ways to pick k people from n, and divide by the k! variants”. Writing this out, we get our combination formula, or the number of ways to combine k items from a set of n:

Sometimes C(n,k) is written as:

which is the the binomial coefficient.

A few examples

Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters).

• Combination: Picking a team of 3 people from a group of 10. $C(10,3) = 10!/(7! * 3!) = 10 * 9 * 8 / (3 * 2 * 1) = 120$.

  Permutation: Picking a President, VP and Waterboy from a group of 10. $P(10,3) = 10!/7! = 10 * 9 * 8 = 720$.

• Combination: Choosing 3 desserts from a menu of 10. C(10,3) = 120.

  Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.

Don’t memorize the formulas, understand why they work. Combinations sound simpler than permutations, and they are. You have fewer combinations than permutations.