Combinatorial identities | AMC Advanced topics (AMC 12)

This chapter only applies to AMC 12 test takers.

This chapter covers a few combinatorial identities that show up on AMC 12 problems, especially in polynomial expansions and counting arguments.

Binomial theorem

The binomial theorem gives a fast, organized way to expand a binomial raised to a power. Instead of multiplying $(a + b)$ by itself repeatedly, you can write down every term and its coefficient in one formula. It’s also useful for counting how many distinct terms appear in an expansion.

$(a + b)^{n} = s u m_{k = 0}^{n} (k n) a^{n - k} b^{k}$

Here’s the key idea for counting: each value of $k$ produces a different pair of exponents $(n - k, k)$ , so it produces a different term $a^{n - k} b^{k}$ .

Example: counting distinct “groups”

Suppose you’re choosing a group of four colors from two colors, $a$ and $b$ , where:

repetition is allowed
order doesn’t matter

A group like “three $a$ ’s and one $b$ ” corresponds to the term $a^{3} b$ . Modeling this with $(a + b)^{4}$ lists all possible exponent pairs.

$(a + b)^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$

There are $5$ distinct terms in the sum, so there are $5$ distinct groups:

$a^{4}$
$a^{3} b$
$a^{2} b^{2}$
$a b^{3}$
$b^{4}$

The coefficients ( $4, 6, 4$ ) count how many ordered selections lead to the same unordered group. For “how many distinct groups?”, you ignore coefficients and count distinct terms.

We can also see the “ $n + 1$ distinct terms” pattern directly from the summation:

$(a + b)^{4} = k = 0 \sum 4 (k n) a^{4 - k} b^{k}$

Since $k$ runs from $0$ to $4$ , there are $5$ possible values of $k$ , and therefore $5$ distinct exponent pairs.

On the other hand, if a question asks for the number of ways to choose exactly $3$ of $a$ and $1$ of $b$ with order counted, then you do want the coefficient of $a^{3} b$ , which is $4$ .

Example: The question below is from 2021 AMC 12A

What is the number of terms with rational coefficients among the $1001$ terms in the expansion of $(x 32 + y 3)^{1000} ?$ A. $0$
B. $166$
C. $167$
D. $500$
E. $501$

(spoiler)

Answer: C. $167$

Pascal’s identity

Pascal’s identity breaks one combination into a sum of two nearby combinations. It’s often used to simplify counting arguments or to connect a problem to Pascal’s triangle.

$(k n) = (k n - 1) + (k - 1 n - 1)$

Pascal’s triangle

Pascal’s triangle is a visual way to organize binomial coefficients.

The outside edges are all $1$ ’s.
Every interior entry is the sum of the two entries diagonally above it.

To find $(4 5)$ , go to the row corresponding to $n = 5$ and the position corresponding to $k = 4$ . In the diagram below, $(4 5)$ is circled.

Notice how the $4$ and $1$ above the circled $5$ add to $5$ . That’s Pascal’s identity in action: each entry equals the sum of the two “parents” above it.

You may also notice another pattern: the sum of the entries in a row is twice the sum of the entries in the previous row. This comes from repeatedly applying Pascal’s identity across the row.

Symmetry

Pascal’s triangle is symmetric: entries mirror across the center. That symmetry corresponds to another useful identity.

$(k n) = (n - k n)$

This says choosing $k$ items from $n$ is equivalent to choosing the $n - k$ items you don’t take.

For example, since $(4 5)$ is $5$ , we also have:

$(5 - 4 5) = (1 5) = 5$

You can see both $5$ ’s in Pascal’s triangle: they appear symmetrically in the same row.

Common themes

Expect combinatorial identity questions to be disguised as questions about paths on a grid, choosing team members, or the number of terms in a polynomial.
Look out for the “at least” constraint. This likely means that there is some simpler way to solve for the complement, or opposite, of what is desired. This will be far easier to find using combinatorial identities. The answer will be the total combinations minus the complement of what is desired. This is called “reverse counting”.