More Than One Way to Solve It, Again

photo by Annie Pilon via flickr

We continue with our counting lessons — and once again, Kitten proves that she doesn’t think the same way I do. In fact, her solution is so elegant that I think she could have a future as a mathematician. After all, every aspiring novelist needs a day job, right?

If only I could get her to give up the idea that she hates math…

Permutations with Complications

How many of the possible distinct arrangements of 1-6 have 1 to the left of 2?

Competition Math for Middle School, by J. Batterson

My Method: Hit the Problem with a Sledgehammer

It took some time, but I did get the answer. I used casework — that is, I thought of all the possible things that might happen to satisfy the requirements, counted them individually, and added up my possibilities to get the final answer.

The cases I considered were all the places that the 1 could go and still have the 2 to its right. Making a 6-digit number, we could have:

1 _ _ _ _ _
_ 1 _ _ _ _
_ _ 1 _ _ _
_ _ _ 1 _ _
_ _ _ _ 1 2

Then I filled in the number of choices available for each blank space, using the Fundamental Counting Principle and taking into account that the 2 has to come after the 1. After that, I multiplied everything out (five multiplications, one for each line). And finally, I added those products together to find my answer.

Ugly, but it works.

The Book’s Method: Flash of Insight Required

I didn’t like the solution in the book because it requires a flash of insight. Such “Aha!” moments are fun but notoriously unreliable. The book used symmetry to solve the problem, recognizing that every possible arrangement of digits, such as:

1 2 3 4 5 6

has a mirror-image partner:

6 5 4 3 2 1

And in every such mirror-image pair, only one of them will have the 1 and 2 in the correct order. So we can just count the number of possible arrangements of digits and then divide by 2.

Nice, if you happen to think of it.

Kitten’s Method: Build the Number, Counting As You Go

Kitten’s method only took two short lines on her whiteboard — though it will take much longer here, since I have to type out her explanation. To begin with, the 1 and 2 must be in this order:

1 2

Choices = 1. Total possible numbers so far:

1 \times  (the choices for the other digits)

I almost interrupted her to say that the 1 and 2 don’t have to sit next to each other. Thankfully, I kept my mouth shut, and thus saved myself from sticking my foot in it. She had everything under control.

The next digit can fit before the 1, or after the 2, or in between them:

_ 1 _ 2 _

Choices = 3. Total possible numbers so far:

1 \times  3 \times  (the choices for the other digits)

Kitten put a 3 at the end for demonstration purposes, but as she explained, the possibilities are the same no matter which digit we use or where we put it.

The next digit has 4 choices of where to go:

_ 1 _ 2 _ 3 _

1 \times  3 \times  4 \times  (the choices for the other digits)

The next digit has 5 choices of where to go:

_ 1 _ 2 _ 3 _ 4 _

1 \times  3 \times  4 \times  5 \times  (the choices for the last digit)

The final digit has 6 choices of where to go:

_ 1 _ 2 _ 3 _ 4 _ 5 _

Total possible numbers that fit our criterion:

1 \times  3 \times  4 \times  5 \times  6


Free-Learning-Guide-Booklets2Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.


10 thoughts on “More Than One Way to Solve It, Again

  1. I wouldn’t describe the symmetry argument as a flash of insight. I always look for symmetries before I start calculating, and it usually pays off. In this case, the simplest symmetry operation is to swap the 1 and the 2, although the book’s idea of reversing the entire permutation also works.

    1. Dave, it’s probably my inexperience (I’ve only learned to count in the last few years) that made the book’s answer seem like a “trick.” I’ve mentioned before that I’m not really good at math.

      Your point about switching the 1 and 2 makes more sense to me — especially since our lesson had been on correcting for the over-counting that results from duplicate letters. So switching two numbers (while leaving everything else alone) would be a logical extension, not like such a huge leap as the book’s answer seemed to me.

  2. Very enlightening piece.Indeed kids do come up with alternate, much more efficient solving methods- in such situations I learn and benefit from understanding their thought processes. Peace.

    (Wow I didn’t know you have such a brilliant daughter; you must be so proud of her.)

  3. Sometimes, you will be surprised that students have solutions that you can never think of. I remember one problem where I have created 19 solutions. Still, there are solutions done by students that was not in my list.

    Excellent work Denise.

  4. Thank you, Mr. Koh and Guillermo. I tell my students that in math, there is almost always more than one way to solve a problem. So why do I get surprised when they prove that’s true?

    In addition to many solutions, sometimes one can create many different problems from the same situation. I really enjoyed an exercise proposed by George Lechner, which became my blog post Puzzle: Random Blocks. Jonathan has a good article about this, using a geometry question: Puzzles: How to modify problems – an example.

  5. Kitten’s method (minus the 1-to-the-left-of-2 requirement) is exactly how I explain permutations on n items to my students. It’s a really illustrative way to show why n! equals the product of the positive integers from 1 to n. It also helps with explaining why 0! = 1.

  6. Good point, Eric. If we teach factorial as a separate, abstract notation (“Multiply the countdown”), then 0! makes no sense. But when taught in the context of permutations, it’s perfectly reasonable. There’s exactly one way you can arrange zero things — that is, by having nothing.

  7. A similar idea can be extended to the concept of combination as well: it is noted that nC0=1
    (understood as n choose zero which is one of the binomial series expansion coefficients).
    I ask my students how many ways can you choose zero objects from n objects? They all reply with a resounding one way, ie that 1 way is simply to choose nothing. Peace.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s