Combinations
Master combinations for Year 11 Specialist Mathematics in Queensland (QCAA). A combination counts the ways to choose objects from a larger set when the order does not matter, written nCr and found from “n choose r”.
You will learn to evaluate nCr from the factorial definition, tell combinations apart from permutations, and use the symmetry rule to solve committee, handshake and geometry problems — the selection foundation for probability and the binomial theorem later in the course.
Theory
A combination counts the number of ways to choose \(r\) objects from \(n\) distinct objects when order does not matter. In Year 11 Specialist Mathematics (QCAA, Queensland) it is written \({}^{n}C_{r}\) or \(\binom{n}{r}\) and found from \({}^{n}C_{r}=\dfrac{n!}{r!\,(n-r)!}\). This page shows the formula, the permutations-vs-combinations distinction, and worked examples.
A combination is an unordered selection of objects from a larger set. Choosing which objects to include is all that matters — the order in which you pick them does not. The selection \(\{A,B\}\) is the same combination as \(\{B,A\}\).
The number of ways to choose \(r\) objects from \(n\) distinct objects is written \({}^{n}C_{r}\), \(\binom{n}{r}\) or \(C(n,r)\), and read "\(n\) choose \(r\)". Here \(n\) is the size of the whole set and \(r\) is the number selected, with \(0\le r\le n\).
The key contrast is with a permutation, which is an ordered arrangement. A committee of two, a doubles pair and a handful of lottery numbers are combinations (order ignored); a president-then-secretary, a race podium and a PIN are permutations (order matters). Ask "does swapping two of the chosen objects give a genuinely different outcome?" — if no, use a combination.
Because each unordered selection of \(r\) objects can be arranged in \(r!\) different orders, permutations always outnumber combinations by a factor of \(r!\): \({}^{n}P_{r}={}^{n}C_{r}\times r!\).
The number of ways to choose \(r\) objects from \(n\) distinct objects, order not important, is:
In practice, cancel the larger factorial in the denominator and keep only \(r\) descending factors on top:
Combinations relate to permutations (ordered) by:
How to evaluate a combination
- Check order does not matter: confirm the selection is unordered (a group, team or pair), so a combination — not a permutation — is required.
- Write the definition: substitute \(n\) and \(r\) into \({}^{n}C_{r}=\dfrac{n!}{r!\,(n-r)!}\).
- Cancel the factorials: divide out \((n-r)!\), leaving \(r\) descending factors over \(r!\).
- Do the arithmetic: multiply the top, multiply the bottom, then divide to a single whole number. Use \({}^{n}C_{r}={}^{n}C_{n-r}\) to keep the numbers small.
Write the definition first, then cancel the \(5!\), then do the arithmetic:
| \({}^{8}C_{3}\) | \(=\) | \(\dfrac{8!}{3!\,(8-3)!}\) |
| \(=\) | \(\dfrac{8!}{3!\,5!}\) | |
| \(=\) | \(\dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}\) | |
| \(=\) | \(\dfrac{336}{6}\) | |
| \(=\) | \(56\) |
\({}^{8}C_{3}=56\).
Order is ignored, so it is a combination \({}^{7}C_{3}\). Write the definition, cancel, then evaluate:
| \({}^{7}C_{3}\) | \(=\) | \(\dfrac{7!}{3!\,(7-3)!}\) |
| \(=\) | \(\dfrac{7!}{3!\,4!}\) | |
| \(=\) | \(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}\) | |
| \(=\) | \(\dfrac{210}{6}\) | |
| \(=\) | \(35\) |
There are \(35\) ways to choose the group.
Each handshake is an unordered pair of people, so count \({}^{8}C_{2}\). Write the definition, cancel the \(6!\), then evaluate:
| \({}^{8}C_{2}\) | \(=\) | \(\dfrac{8!}{2!\,(8-2)!}\) |
| \(=\) | \(\dfrac{8!}{2!\,6!}\) | |
| \(=\) | \(\dfrac{8 \times 7}{2 \times 1}\) | |
| \(=\) | \(\dfrac{56}{2}\) | |
| \(=\) | \(28\) |
There are \(28\) handshakes.
(a) A team is an unordered selection, so use the combination \({}^{7}C_{3}\):
| \({}^{7}C_{3}\) | \(=\) | \(\dfrac{7!}{3!\,(7-3)!}\) |
| \(=\) | \(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}\) | |
| \(=\) | \(\dfrac{210}{6}\) | |
| \(=\) | \(35\) |
(b) The three roles are different, so order matters — use the permutation \({}^{7}P_{3}\):
| \({}^{7}P_{3}\) | \(=\) | \(7 \times 6 \times 5\) |
| \(=\) | \(210\) |
Relate the two counts: each team of \(3\) can be arranged into roles in \(3!\) ways:
| \({}^{7}P_{3}\) | \(=\) | \({}^{7}C_{3} \times 3!\) |
| \(210\) | \(=\) | \(35 \times 6\) |
(a) \(35\) teams. \;(b) \(210\) role assignments. The \(210\) arrangements are \(3!=6\) times the \(35\) selections, because order multiplies each selection by \(3!\).
Common pitfalls
Frequently asked questions
What is the formula for a combination nCr?
The number of ways to choose \(r\) objects from \(n\) distinct objects, order not important, is \({}^{n}C_{r}=\dfrac{n!}{r!\,(n-r)!}\).
What is the difference between a permutation and a combination?
A permutation is an ordered arrangement, so \(AB\) and \(BA\) are different. A combination is an unordered selection, so \(\{A,B\}\) counts once. They are linked by \({}^{n}P_{r}={}^{n}C_{r}\times r!\).
How do you know whether to use a combination or a permutation?
Ask whether swapping two chosen objects gives a different outcome. If order does not matter (a team, group or pair), use a combination; if order matters (roles, a podium, a PIN), use a permutation.
Why does nCr equal nC(n-r)?
Choosing the \(r\) objects to include automatically fixes the \(n-r\) objects left out, so the two selections pair up one-to-one. Hence \({}^{n}C_{r}={}^{n}C_{n-r}\); for example \({}^{7}C_{2}={}^{7}C_{5}=21\).
How many handshakes occur if everyone shakes hands once?
Each handshake is an unordered pair of people, so for \(n\) people the number of handshakes is \({}^{n}C_{2}=\dfrac{n(n-1)}{2}\). For \(8\) people that is \(28\).
What are nC0 and nCn equal to?
Both equal \(1\): there is exactly one way to choose none of the objects and one way to choose all of them. This matches the formula since \(0!=1\).