The same question, twice

What do you think?

You have 6 people and need to pick 2 for a team. Is C(6,2) (choosing 2 to be ON the team) equal to C(6,4) (choosing 4 to be OFF)?

Try it yourself, watch how both groups change together.

Pascal's Triangle

Team Selector

On Team (0)

Off Team (6)

C(6,0)

C(6,6)

Why counting both ways proves something

Each selection defines both groups at once. This gives us:

Symmetry Identity

\binom{6}{2} = 15

There are 15 ways to choose 2 people from 6 for the team.

Step 1 of 4

What does C(10,7) equal using the symmetry identity? (e.g. C(5,2))

1/2

A story proof works by finding a counting question, then counting the same thing two different ways. If both methods count the same set, the answers are equal.

Vandermonde's identity

What do you think?

You have 5 work friends and 4 neighbors. You invite 3 people total. How many ways can you do this?

But there's another way to count the same thing.

Dinner Party Planner

From work: 0

Work (4)

C(4,0) = 1

Neighbors (3)

C(3,3) = 1

0 from work + 3 neighbors = 3 guests

1 × 1 = 1 ways

All splits

1 + 12 + 18 + 4 = 35

= C(7,3)

What do you think?

Same scenario: 5 work friends, 4 neighbors, invite 3 total. You could invite (2 work, 1 neighbor) or (1 work, 2 neighbors) or (0 work, 3 neighbors) or (3 work, 0 neighbors). What's the total?

Enter a whole number

Both methods count the same dinner parties:

Vandermonde's Identity

\binom{m+n}{k}

Left side: Pick k people from the combined pool of m+n friends.

Step 1 of 3

Using Vandermonde: C(7,3) = C(4,0)×C(3,3) + C(4,1)×C(3,2) + C(4,2)×C(3,1) + C(4,3)×C(3,0). Calculate the right side. (whole number)

The hockey stick

One more identity:

$\Large\binom{n}{0} + \binom{n+1}{1} + \binom{n+2}{2} + \cdots + \binom{n+k}{k} = \binom{n+k+1}{k}$

It's called the hockey stick because of how it looks on Pascal's triangle.

What do you think?

Calculate: C(5,0) + C(6,1) + C(7,2) + C(8,3). What pattern do you notice?

Enter a whole number

Picture a staircase where you make exactly $k$ upward climbs.

Staircase Paths

Position of last climb

Last climb at step 5

C(4,2) = 6 paths

Sum over all positions

13610

1 + 3 + 6 + 10 = 20

= C(6,3)

What do you think?

How many paths to point (n+k, k) if you can only move right or up?

The method

Story proofs aren't tricks, they're how combinatorialists think.

The Story Proof Method

\text{Find a counting question}

What are these expressions counting? People on a team? Paths on a grid? Ways to invite friends?

Step 1 of 3

Try a story proof when you see binomial coefficients on both sides of an equation. Ask yourself: what are these expressions counting? Is there a single scenario where both sides naturally appear?

Test your understanding

Using symmetry: C(100,97) = C(100,?) (whole number)

True or False: C(10,3) + C(10,7) = C(11,3) using Vandermonde with m=10, n=0. (true or false)

Hockey stick: C(4,0) + C(5,1) + C(6,2) = C(?,2). What replaces the ? (whole number)

Calculate C(50,48) the easy way. (whole number)

Using Vandermonde: C(8,3) where we split 8 into 5+3. One term is C(5,2)×C(3,1). How many terms total in the sum? (whole number)