Factorials

The product $n\times(n-1)\times\cdots\times1$ that counts the arrangements of $n$ distinct objects comes up so often that it gets its own name and symbol: $n!$, read "$n$ factorial." This lesson defines the factorial, shows how to simplify factorial expressions without multiplying everything out, and covers the special convention $0!=1$.

What you will learn

• Define n! and compute small factorials
• Simplify ratios and products of factorials
• Recognize factorial growth and common factorial identities

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

Defining the factorial

For a positive whole number $n$, the factorial $n!$ is the product of every whole number from $n$ down to $1$: $n!=n\times(n-1)\times(n-2)\times\cdots\times2\times1$. So $5!=5\times4\times3\times2\times1=120$. Factorials grow extremely quickly — $10!$ is already over three million — so contest problems almost never expect you to compute a large factorial fully; instead they expect you to simplify a ratio or expression before multiplying.

By convention, $0!=1$. This might look strange at first, since there is no product of zero terms to multiply, but the convention exists so that formulas involving factorials — including the arrangement-counting formulas used throughout this course — continue to work correctly at the boundary case of arranging zero objects. Treat $0!=1$ as a fact to memorize rather than something to re-derive each time.

Simplifying factorial ratios

A ratio like $\dfrac{8!}{6!}$ should almost never be computed by finding $8!$ and $6!$ separately and dividing. Instead, write out the larger factorial only as far as the smaller one, and cancel: $\dfrac{8!}{6!}=\dfrac{8\times7\times6\times5\times4\times3\times2\times1}{6\times5\times4\times3\times2\times1}=8\times7=56$. Every factor from $6$ down to $1$ appears in both the numerator and the denominator and cancels completely, leaving only the factors above $6$ in the numerator.

This shortcut works whenever one factorial is a ratio against a smaller factorial: $\dfrac{n!}{k!}$ simplifies to the product of the whole numbers from $n$ down to $k+1$, with $n-k$ factors in total. Recognizing this pattern turns a seemingly large computation into a short multiplication, and it is the single most useful factorial skill for contest problems.

Factorials built from smaller factorials

Every factorial can be broken into a single factor times a smaller factorial: $n!=n\times(n-1)!$. This identity is useful for simplifying expressions that mix factorials of different sizes, such as $\dfrac{n!}{(n-1)!}=n$ directly from the definition, without cancelling term by term.

When an expression contains a factorial in both the numerator and the denominator with more complicated numbers attached, look for the largest factorial that divides evenly into the others first, and expand only as much as needed. This keeps the arithmetic small and makes it far less likely that a term gets miscancelled or dropped.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Computing both factorials in a ratio completely instead of cancelling shared factors first.
• Forgetting the convention $0!=1$ and treating $0!$ as $0$.
• Dropping or miscounting a factor when expanding a factorial only partway.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• $n!$ is the product of every whole number from $n$ down to 1, and $0!=1$ by convention.
• Simplify a factorial ratio by expanding the larger factorial only far enough to cancel the smaller one.
• Break a factorial into $n\times(n-1)!$ to relate factorials of different sizes without full expansion.