Selections
Master selections, also called combinations, for Queensland Year 11 Mathematical Methods (QCAA). A selection counts the ways to choose a group where the order does not matter, such as picking a committee, team or hand of cards.
You will learn to count combinations, handle restricted selections by grouping and multiplying, solve "at least" problems with the complement, and connect selections to Pascal's triangle — skills that underpin probability.
Every question with a fully worked solution.
- Selections - Video - Selections Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a selection is an unordered choice of objects, counted with the combination \(^{n}C_{r}=\dbinom{n}{r}=\dfrac{n!}{r!\,(n-r)!}\). This page shows how to count committees, teams and hands where order does not matter, selections with a restriction (split into groups and multiply), at-least problems, and the link between combinations and Pascal's triangle.
A combination is a selection of \(r\) objects from \(n\) distinct objects where order does not matter. It is written \(^{n}C_{r}\), \(\dbinom{n}{r}\) or 'n choose r', and equals \(\dfrac{n!}{r!\,(n-r)!}\).
Because order is ignored, a combination divides the ordered count \(^{n}P_{r}\) by the \(r!\) rearrangements of the chosen objects. A committee, a team and a hand of cards are all selections.
Combinations obey the symmetry \(^{n}C_{r}={}^{n}C_{n-r}\) and are exactly the entries of Pascal's triangle: row \(n\) reads \(^{n}C_{0},{}^{n}C_{1},\dots,{}^{n}C_{n}\).
Combination — choose \(r\) from \(n\), order ignored:
Symmetry of combinations:
How to count a selection
- Check order: confirm order does not matter, so it is a combination \(^{n}C_{r}\), not a permutation.
- Apply restrictions: for must-include or must-exclude items, fix or remove them first; for groups, choose from each group and multiply.
- At least / at most: add the separate cases, or count the total and subtract the unwanted selections (the complement).
Use \(^{n}C_{r}=\dfrac{n!}{r!\,(n-r)!}\) with \(n=8,\ r=3\):
| \(^{8}C_{3}\) | \(=\) | \(\dfrac{8!}{3!\,5!}\) |
Cancel the \(5!\) and expand \(3!\):
| \(=\) | \(\dfrac{8\times 7\times 6}{3\times 2\times 1}\) | |
| \(=\) | \(\dfrac{336}{6}\) |
Divide:
| \(^{8}C_{3}\) | \(=\) | \(56\) |
There are \(56\) selections.
Order does not matter, so choose \(4\) from \(10\) with \(^{10}C_{4}\):
| \(^{10}C_{4}\) | \(=\) | \(\dfrac{10!}{4!\,6!}\) |
Cancel the \(6!\) and expand \(4!\):
| \(=\) | \(\dfrac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\) | |
| \(=\) | \(\dfrac{5040}{24}\) |
Divide:
| \(^{10}C_{4}\) | \(=\) | \(210\) |
There are \(210\) committees.
Split into groups — choose \(2\) of the \(4\) women and \(1\) of the \(5\) men:
| \(\text{women}\) | \(=\) | \(^{4}C_{2}=\dfrac{4\times 3}{2\times 1}=6\) |
| \(\text{men}\) | \(=\) | \(^{5}C_{1}=5\) |
Multiply the two group counts (multiplication principle):
| \(N\) | \(=\) | \(6\times 5\) |
| \(=\) | \(30\) |
There are \(30\) committees.
Count the complement — all teams, minus teams with no girl (all \(3\) from the \(5\) boys):
| \(\text{total}\) | \(=\) | \(^{8}C_{3}=\dfrac{8\times 7\times 6}{6}=56\) |
| \(\text{no girl}\) | \(=\) | \(^{5}C_{3}=\dfrac{5\times 4\times 3}{6}=10\) |
Subtract the unwanted teams:
| \(N\) | \(=\) | \(56-10\) |
| \(=\) | \(46\) |
There are \(46\) teams with at least one girl.
Common pitfalls
Frequently asked questions
When do I use a combination instead of a permutation?
Use a combination \(^{n}C_{r}\) when order does not matter (a team, a committee, a hand); use a permutation when order matters.
What is the formula for nCr?
\(^{n}C_{r}=\dfrac{n!}{r!\,(n-r)!}\). It divides the ordered count \(^{n}P_{r}\) by \(r!\) because the \(r\) chosen objects can be reordered.
How are combinations linked to Pascal's triangle?
The entries of row \(n\) of Pascal's triangle are \(^{n}C_{0},{}^{n}C_{1},\dots,{}^{n}C_{n}\); for example row \(4\) is \(1,4,6,4,1\).
How do I count a selection with two groups?
Choose the required number from each group separately, then multiply the counts together using the multiplication principle.
What is the quickest way to do at least one?
Count the total selections and subtract the ones with none of the required type: \(N=\text{total}-\text{none}\).