Posted in Combinatorics

What makes counting problems difficult

I don’t have the numbers but I think not many will disagree with me that among those who like mathematics and find joy in the challenge of solving math problems, only very few will also like to solve combinatorial or counting problems. And if I may be allowed to push my observation a bit further, those who are more inclined to algebra are the ones who have more aversion to problems that asks them to count.

Indeed, combinatorial problems are odd. Unlike other mathematics problems, these type of problems cannot easily be categorized and solved with predictable algorithms. Each problem always seems to be a case in itself. Knowing all the formulas for different cases of permutations and combinations is not a guarantee that one will be able to solve these problems. For many, these are only useful if the problem already explicitly asks for number of permutations and combinations. And when indeed the problem says so, one still has to decipher other conditions and the assumptions embedded in the context. Knowing when to add or multiply is also difficult, if not the most difficult part. Knowing when one already double counted also requires depth of understanding not only of the formulas but also of the context of the problem. Of course for some these are also the very reason why they love to solve counting problems.

Take the case of this family of problems:

 

  1. How many three-digit numbers can you form from the digits, 5, 6, 7, 8, and 9? (This requires direct application of the permutation formula or better, the multiplication principle.)
  2. How many three-digit numbers are there in all? (That you will choose from 0 to 9 is implied only plus you ask yourself how about 027? Is it a 3-digit number of a 2-digit number?)
  3. How many numbers can be formed from 5, 6, 7, 8, and 9?
  4. How many three-digit numbers greater that 600 can be formed from the digits 5, 6, 7, 8, and 9? (Oops … so what formula is applicable now?)
  5. How many numbers greater than 3,000,000 can be formed by arrangements of the digits 1, 2, 2, 3, 6, 6, 6? (More Ooops…)
Having identified some of the difficulties inherent to counting problems, what teaching tip can I propose?
  1. Ask for different ways of solving the problem, slowly discouraging students’ reliance to listing. Thinking in terms of tree diagrams and empty boxes are great help.
  2. Defer introduction of formulas before students can solve the problems like the ones above using the fundamental counting principle. Permutations and combination formulas are only powerful to the extent to which students understand the multiplication principle.
  3. Another pedagogical tip is to ask students to identify what makes a problem similar and different in structure from the problems they previously solved. If they can specify which problem, the better.
  4. As a consequence of tip#3, problems are best given per family with family defined in terms context (e.g., the family of problems given in the example above) and not in terms of similarity in solution structure.

Please add yours. Thanks.

You may want to read my two other posts on combinatorics:

Some useful references:
Posted in Combinatorics

Pascal’s triangle and Counting Permutations

This is the second in my series of posts in combinatorics. The first post links the Fundamental  Counting Principle, Powers of 2, and the Pascal Triangle. This second post connects the Pascal’s Triangle and the formula for counting the number of permutations with identical objects. The context for connections is a puzzle about counting the total routes of a rook to squares in a chessboard.

The Puzzle

Trace the shortest route the rook can traverse from its corner position to the opposite corner. How many such routes are there?

Solution 1

A useful problem solving strategy here is to simplify the problem: Count the number of distinct paths the rook will land to the nearest square.

To determine the shortest path, the rook should either go north or east or left or right only. That is, the rook has only two choices each time it moves through a square as shown in the figure below. You will not reach the third or fourth line before you figure out that the pattern generates the Pascal’s Triangle.

Thus there are 3432 distinct shortest paths the rook can traverse from one corner to the opposite corner.

Solution 2

Another way to solve puzzle 2 is shown below. The arrows trace some possible shortest paths for the rook.

The paths of the arrows in the figure can be represented this way: (N = North, E = East)

  1. N N N E N E E N E N E N E E
  2. E E E E N N N E E N N N N E
  3. E E E E E E E N N N N N N N
Each shortest path from one corner square to the opposite corner square is made up of 7 N’s and 7 E’s. Thus, counting the total number of shortest paths is the same as counting the total number of distinct arrangements of 7 N’s and 7 E’s in a row. The formula for counting the permutation of n objects a, b, c, … of which are identical will be useful here. This is given by the formula: P = \frac{n!}{a!b!c!...}. So,
P = \frac{14!}{7!7!} = 3432.
Click here to see solutions to this problem using the multiplication principle and combination formula.
If you want to practice your skill in solving permutation and combination problems, Examrace Permutation & Combination Made Easy (Examrace Aptitude Series) will be useful.
Posted in Combinatorics

The Counting Principle, Pascal’s Triangle, and Powers of 2

This post shows how we can help students make connections among counting principle, the Pascal’s triangle, and powers of 2. I have tried this lesson in an in-service training program but I’ve yet to test it with students in high school. The lesson uses the strategy Teaching thru Problem Solving.

A piece of knowledge is powerful to the extent to which it is connected to other piece of knowledge. The more connections there are, the more powerful it becomes. Mathematics teaching therefore should always aim to help students make connections among the different concepts of mathematics. You may want to read  my article about   understanding as making connections.

The Problem: Trace the paths that will spell “MATHEMATICS” starting from the letter M on top moving only downwards, either to the immediate letter to its right or to the immediate letter to its left. How many different paths are there in all?

After a few minutes and the class is seem getting nowhere you may suggest to students to try simpler case first  like trying the word MATH. Trying simpler case is a good problem solving strategy and habit students need to learn.

Solution 1

Suppose we  spell the word “MATH” only. From M we can move downwards and may either choose the A at the left or the A at the right. Having chosen an A we can either choose the T down left or the T down right. And having chosen one we can either choose the H down right or the H down left. Each time we only have two choices. Thus, the number of ways of tracing the word “MATH” in the above figure is

2·2·2 =23=8

Using the same line of thinking, the total number of paths which spells “MATHEMATICS” is

 2·2·2·2·2·2·2·2·2·2=210=1024.

Solution 2

Notice the number of arrows that converges to a particular letter. It tells the number of paths that pass through it. Thus, to count the number of ways of tracing the word “MATH” we only have to add the total number of arrows that point to the H’s. There are

 1 + 3 + 3 + 1 = 8.

Count the number of arrows converging to each letter in MATHEMATICS . You will generate the triangular array of numbers below.

The number of arrows converging to S is

1 + 10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 1024 or  210.

The solutions showed two important principles of counting.

The Multiplication Principle. If one task can be done in m ways and then another task can be done in n ways, the pair of tasks, first one and then the other, can be performed in

m n ways.

 The Addition Principle. If one task can be done in m ways and another task in n ways, then one task or the other can be done in

m + n ways.

Anyone who wants to understand permutations, combinations and anything that involves counting should first understand these principles.

The triangular array of numbers generated above is one of the most influential number patterns in the history of mathematics. It is called Pascal’s triangle after the renowned French mathematician Blaise Pascal (1623-1662) who discovered it. The triangle is also called Yang Hui’s triangle in China as the Chinese mathematician Yang Hui discovered it much earlier in 1261. The same triangle was also in the book “Precious Mirror of the Four Elements” by another Chinese mathematician Chu-Shih-Chieh in 1303.

The Pascal triangle yields interesting patterns and relationships. Some of the obvious ones are:

  1. To generate the next row, you will have to add the two numbers above it.
  2. Another striking property of this array of numbers is its symmetry. Note the numbers on both sides of the middle number in each row.
  3. The sum of the numbers in each row can be expressed in powers of two.

Recommended readings on combinatorics:

  1. Mathematics of Choice: Or, How to Count Without Counting (New Mathematical Library)
  2. Counting: The Art of Enumerative Combinatorics (Undergraduate Texts in Mathematics)
  3. Introductory Combinatorics (5th Edition)