How Do You Calculate nCr (Combinations)?

Calculate combinations using $\binom{n}{r}=\dfrac{n!}{r!(n-r)!}$, where $n$ is the total number of objects and $r$ is the number selected. Combinations apply when the order of selection does not matter.

How the formula works

The notation $\binom{n}{r}$, also written as ${}^{n}C_r$ or (nCr), means “the number of ways to choose objects from objects.” The symbol (!) represents a , so , with .

The formula begins with all ordered arrangements of $r$ selected objects, then divides by (r!) because each chosen group can be ordered in (r!) equivalent ways. This explains why combinations ignore order rather than merely listing selections.

For example, to choose $2$ students from a group of $5$:

$$\binom{5}{2}=\frac{5!}{2!3!} =\frac{5\times4\times3!}{2\times1\times3!} =\frac{20}{2}=10.$$

Therefore, there are $10$ possible groups. Cancelling common factorials before multiplying makes the calculation faster and reduces arithmetic errors.

Before calculating, identify the decision being made and whether swapping two selected objects creates a new outcome. In choosing $3$ books from $8$, swapping the order does not create a new set, so $\binom{8}{3}=56$. If the same three books were awarded first, second and third prizes, the outcomes would be ordered and combinations alone would be incorrect.

The values must satisfy $0\leq r\leq n$, with $n,r\in\mathbb{N}$. A useful symmetry property is

$$\binom{n}{r}=\binom{n}{n-r}.$$

For example, $\binom{6}{4}=\binom{6}{2}=15$. On a GDC, enter , select the combinations or (nCr) command, and then enter .

In IB Mathematics: Analysis and Approaches SL 1.9, combinations appear as coefficients in the binomial theorem:

$$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r.$$

Exam technique

On Paper 1, show the factorial formula and simplification because calculators are not permitted. On Paper 2, a GDC may calculate (nCr), but show how the values of $n$ and $r$ arise.

For a Find question, include relevant stages of working; the command term requires an answer with the relevant method shown, not only a calculator result.

A common misconception is to use combinations when order matters. Choosing two committee members uses combinations, but assigning a president and vice-president does not, because the roles make the order significant.