Factorial Notation & nCr
Pascal's triangle is quick for small rows and painful for large ones. Factorials give a formula that reaches any entry of any row directly — no triangle required.
Factorial notation
For a positive integer $n$, the factorial $n!$ ("$n$ factorial") is the product of every whole number from $n$ down to $1$:
$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$
So $5! = 5\times4\times3\times2\times1 = 120$, and $3! = 3\times2\times1=6$. By definition, $0! = 1$ — not $0$. This isn't arbitrary: it makes every formula built from factorials behave correctly at the edges, as you'll see below.
Factorials often appear as a ratio, and most of the terms cancel. For instance $\dfrac{7!}{5!} = \dfrac{7\times6\times5\times4\times3\times2\times1}{5\times4\times3\times2\times1} = 7\times6=42$ — everything from $5!$ downward cancels, leaving only $7\times6$. Recognising this saves enormous amounts of arithmetic; never expand a factorial fully if it's about to be divided by a smaller one.
The nCr formula
The number $\dbinom{n}{r}$ (read "$n$ choose $r$", also written $^nC_r$) counts the number of ways to choose $r$ objects from a group of $n$, where order doesn't matter. It's defined by:
$\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$
This is exactly the entry at position $r$ in row $n$ of Pascal's triangle. The formula reaches it directly — for row $10$, position $3$, you don't need rows $0$ through $9$; you just compute $\binom{10}{3} = \dfrac{10!}{3!\,7!} = 120$.
$\binom{n}{r}$ counts the ways to choose $r$ items from $n$. Choosing $r$ items is the same as choosing which $r$ of the $n$ positions get a "$b$" when expanding $(a+b)^n$ — which is exactly what each Pascal's-triangle entry represents. The formula and the triangle are two routes to the same count; the formula is just faster once $n$ is large.
Two useful properties
Symmetry: $\dbinom{n}{r} = \dbinom{n}{n-r}$ — choosing which $r$ to include is the same as choosing which $n-r$ to leave out. Edge values: $\dbinom{n}{0} = \dbinom{n}{n} = 1$ always — there's exactly one way to choose nothing, and exactly one way to choose everything. This is also why $0!=1$ is essential: it makes $\binom{n}{0} = \dfrac{n!}{0!\,n!} = 1$ come out correctly rather than being undefined.
Worked example 1
This is exactly $\binom{8}{2}=28$ — the same calculation as "choose 2 from 8", and the same as the third entry of row 8 of Pascal's triangle. Recognising the pattern $\binom{n}{2} = \dfrac{n!}{2!(n-2)!} = \dfrac{n(n-1)}{2}$ is worth memorising, since it appears constantly.
Worked example 2
Symmetry isn't just a nice fact — it's a genuine shortcut. $\dbinom{20}{18}$ looks intimidating, but it equals $\dbinom{20}{2} = \dfrac{20\times19}{2}=190$, which is far quicker. Always check whether $r$ or $n-r$ gives the smaller, easier calculation.
Worked example 3
Quadratics from $\binom{n}{r}$ equations almost always produce one valid root and one that must be rejected on the grounds that $n$ has to be a positive integer (often the equation gives a negative solution). Always state explicitly why the second root is rejected — it's a method mark in most schemes.
Worked example 4
With two different binomial coefficients in the same equation, write both in terms of $n$ using the factorial formula before attempting to solve.
Dividing an equation by a variable is only valid once you know that variable can't be zero — otherwise you risk losing a solution. Here the question states $n>2$, which guarantees $n\neq0$, so the division loses nothing. Had $n=0$ been a genuine possibility, the correct approach would have been to factorise out $n$ and consider both factors separately, exactly as with any other quadratic.
$\binom{n}{r}$ is only defined when $n$ and $r$ are non-negative integers with $r \leq n$. If solving an equation produces a fractional, negative, or otherwise invalid value, it must be rejected — always check your answer makes sense as a genuine "choose" count before finishing.
Check yourself
Evaluate $\dbinom{12}{3}$, and use the symmetry property to state the value of $\dbinom{12}{9}$ without recalculating.
(Hint: cancel the larger factorial first, then use $\binom{n}{r}=\binom{n}{n-r}$.)