5. Pascal’s triangle

In the last lesson, we learned how to count selections when order does not matter.

Now we are going to meet a number pattern that seems almost magical at first — but turns out to be built from exactly the same ideas.

A triangle of numbers

Start with a single 1. Then put 1s down both edges. Every number inside is found by adding the two numbers just above it.

1
1    1
1    2    1
1    3    3    1
1    4    6    4    1
1    5    10    10    5    1

This is called Pascal’s triangle.

Practice puzzle
Continue the triangle
What is the next row?
💡
Learning tipStart and finish with 1. Each inside number is the sum of the two above it.
Reveal answer

1   6   15   20   15   6   1

Why does this belong in a counting course?

Look at the row

1   5   10   10   5   1

These numbers answer every possible “choose from 5” question:

  • choose 0 from 5: 1 way;
  • choose 1 from 5: 5 ways;
  • choose 2 from 5: 10 ways;
  • choose 3 from 5: 10 ways;
  • choose 4 from 5: 5 ways;
  • choose all 5: 1 way.
Main idea
Each row contains the combination numbers

Row n of Pascal’s triangle gives the numbers of ways to choose 0, 1, 2, …, n objects from n objects.

Practice puzzle
Choose a committee

There are 6 students. A committee of 2 is to be chosen.

Use Pascal’s triangle to find the number of possible committees.
💡
Learning tipFind the row for 6 objects, then look at the entry for choosing 2.
Reveal answer

The row for 6 is:

1   6   15   20   15   6   1

The third entry is 15, so there are 15 committees.

Why do the entries add like that?

Suppose we want to choose 2 people from a group of 5, and one of them is Alice.

Every possible committee falls into exactly one of two cases:

  • Alice is chosen;
  • Alice is not chosen.

If Alice is chosen, we choose 1 more person from the remaining 4. If Alice is not chosen, we choose both people from the remaining 4.

4 + 6 = 10

That is exactly the Pascal rule: two neighbouring entries above add to the entry below.

Practice puzzle
Include Ben or not?

Seven children are available and we want a team of 3.

Split the teams into those containing Ben and those not containing Ben. How many are there altogether?
💡
Learning tipIf Ben is included, how many more people must you choose from the other 6?
Reveal answer

With Ben: choose 2 from the other 6, giving 15.

Without Ben: choose 3 from the other 6, giving 20.

15 + 20 = 35

There are 35 teams.

Patterns hiding in the triangle

Pascal’s triangle contains many patterns. Here are three worth noticing.

  • The triangle is symmetric.
  • The first diagonal is all 1s.
  • The next diagonal is 1, 2, 3, 4, 5, …
Practice puzzle
Why is it symmetric?

Why should choosing 2 people from 7 give the same number as choosing 5 people from 7?

Explain this without doing any arithmetic.
💡
Learning tipChoosing 2 people to include is the same decision as choosing which 5 people are left out.
Reveal answer

Every choice of 2 people determines exactly one group of 5 people left over, and vice versa. So the two counts must be equal.

Challenge
Add a whole row

Add the numbers in the first few rows of Pascal’s triangle.

What pattern do you notice? Can you explain why it happens?

What should the sum of row 8 be?
💡
Learning tipA row counts all possible subsets of an n-element set, grouped by their size. For each object there are two choices: in or out.
Reveal answer

The row sums are 1, 2, 4, 8, 16, 32, …

For n objects, every object can either be chosen or not chosen, giving:

2n

So row 8 has sum 28 = 256.

What have we learned?

Pascal’s triangle is not just a pretty number pattern. Its rows encode the numbers of ways to choose objects without ordering.

  • Each inside entry is the sum of the two above it.
  • Row n gives the counts for choosing 0, 1, 2, …, n objects from n.
  • The symmetry reflects choosing what is included or what is left out.
  • A whole row sums to 2n.

Next: Clever counting

Sometimes the quickest way to count something is not to count it directly at all.

In the final lesson we will use complements, cases, symmetry and double counting to solve problems that look much harder than they really are.

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