Binomial Identity

Combination counts are not isolated numbers — they fit together in patterns. This lesson develops two of the most useful patterns: a symmetry that lets a hard combination be rewritten as an easy one, and Pascal's rule, which builds any combination count from two smaller ones without touching a factorial.

What you will learn

• Use the symmetry identity C(n,r) = C(n,n-r)
• Apply Pascal's rule to relate consecutive binomial coefficients
• Recognize binomial coefficients inside counting arguments

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

Binomial coefficients and Pascal's triangle

The values $\binom{n}{r}$ are called binomial coefficients, and arranging them by row, with $n$ increasing downward and $r$ increasing left to right, produces Pascal's triangle. Row $n=4$, for instance, reads $\binom{4}{0},\binom{4}{1},\binom{4}{2},\binom{4}{3},\binom{4}{4}$, which evaluate to $1,4,6,4,1$.

Every entry in this triangle answers a counting question: $\binom{4}{2}=6$ is the number of ways to choose 2 objects from 4, and it is no coincidence that this matches the number of ways to choose the other 2 objects to leave out. The rest of this lesson explains two patterns visible in the triangle: rows are symmetric left to right, and every interior entry is the sum of the two entries diagonally above it.

The symmetry identity

Choosing which $r$ objects out of $n$ to include is exactly the same decision as choosing which $n-r$ objects to leave out — every choice of an "included" group determines a unique "excluded" group, and vice versa. Because these two counting questions describe the same set of decisions, they must have the same answer: $\binom{n}{r}=\binom{n}{n-r}$.

This identity is most useful when $r$ is large and close to $n$, since $\binom{n}{n-r}$ then has a small bottom number and needs far fewer factors to compute. For example, computing $\binom{20}{18}$ directly would require an 18-term factorial ratio, but $\binom{20}{18}=\binom{20}{2}$ needs only two.

Pascal's rule

Pascal's rule states $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$. It comes from casework on a single fixed object, say the last one added to the set of $n$: either that object is included in the chosen group of $r$, or it is not. If it is included, the remaining $r-1$ slots are filled from the other $n-1$ objects, giving $\binom{n-1}{r-1}$ ways. If it is excluded, all $r$ slots are filled from the other $n-1$ objects, giving $\binom{n-1}{r}$ ways. These two cases are disjoint and cover every possibility, so they add.

This rule is the reason each row of Pascal's triangle can be built directly from the row above it by adding adjacent pairs, without ever multiplying out a factorial. It is especially useful when a few nearby combination values are already known and a related one is needed quickly.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Believing Pascal's rule adds two combinations from the same row instead of two adjacent entries in the row above.
• Misapplying the symmetry identity, such as writing $\binom{n}{r}=\binom{n}{r-n}$, instead of the correct $\binom{n}{r}=\binom{n}{n-r}$.
• Recomputing a full factorial ratio by hand instead of noticing that symmetry or Pascal's rule already gives a shortcut.

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

• Pascal's triangle arranges the combination counts $\binom{n}{r}$ by row, one row per value of $n$.
• Symmetry: $\binom{n}{r}=\binom{n}{n-r}$, since choosing which objects to include also decides which to leave out.
• Pascal's rule: $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$, found by casework on whether one particular object is included.