Combinations with restrictions
Combinations with restrictions are unordered selections that must meet a condition, a key skill in Year 11 Specialist Mathematics in Queensland (QCAA) — a member included or excluded, an "at least" or "at most" rule, or a set number from two groups.
You will learn to adjust the pool for a fixed member, split conditions into cases or use the complement, and multiply choices across two groups — selection skills behind committee, team and probability problems.
Theory
Combinations with restrictions count unordered selections that must meet a condition, a core skill in Year 11 Specialist Mathematics (QCAA, Queensland). A member may have to be included or excluded, a selection may need "at least" or "at most" so many of a type, or members may be chosen from two groups. This page shows how to handle each with \({}^{n}C_{r}\).
A combination is an unordered selection: the order in which items are picked does not matter. The number of ways to choose \(r\) items from \(n\) distinct items is written \({}^{n}C_{r}\).
A restriction is an extra condition the selection must satisfy. The commonest are: a particular member must be included, a particular member must be excluded, the selection must contain at least \(k\) or at most \(k\) of a type, or a fixed number must come from each of two separate groups.
Handle an include or exclude rule by adjusting the pool before you count: fixing a member in reduces both the number still to choose and the pool; ruling a member out reduces only the pool.
An "at least" / "at most" rule is a set of allowed cases. Count each case with a product of \({}^{n}C_{r}\) terms, then add the cases — or subtract the unwanted cases from the total (the complement).
The number of unordered selections of \(r\) items from \(n\) distinct items:
If a particular member must be included, fix that member and choose the rest: \(r-1\) more from the \(n-1\) that remain.
If a particular member must be excluded, choose all \(r\) from the \(n-1\) that remain.
To choose \(j\) from a group of \(a\) and \(k\) from a separate group of \(b\), multiply the group choices:
How to count a restricted selection
- Read the restriction: is a member fixed in or out, is it an "at least / at most" condition, or a fixed number from each of two groups?
- Adjust the pool for include/exclude: a fixed-in member drops both the number still to choose and the pool by one; a ruled-out member drops only the pool by one.
- Split "at least / at most" into cases (or use the complement), and for two groups choose from each pool separately.
- Combine: multiply choices made together within a selection; add separate cases. Evaluate each \({}^{n}C_{r}\).
Fix Mia in the team, then work out how many are still to choose and from how many:
| \(\text{team size}\) | \(=\) | \(7\) |
| \(\text{Mia is in, still to choose}\) | \(=\) | \(7-1\) |
| \(=\) | \(6\) | |
| \(\text{players left to choose from}\) | \(=\) | \(10-1\) |
| \(=\) | \(9\) |
Choose the remaining \(6\) from the \(9\) that remain:
| \(\text{ways}\) | \(=\) | \({}^{9}C_{6}\) |
| \(=\) | \(\dfrac{9\times8\times7}{3\times2\times1}\) | |
| \(=\) | \(84\) |
\(84\) ways.
Rule Jack out first — this shrinks only the pool, not the number chosen:
| \(\text{members}\) | \(=\) | \(8\) |
| \(\text{Jack is out, so choose from}\) | \(=\) | \(8-1\) |
| \(=\) | \(7\) | |
| \(\text{still choosing}\) | \(=\) | \(3\) |
Choose all \(3\) from the \(7\) that remain:
| \(\text{ways}\) | \(=\) | \({}^{7}C_{3}\) |
| \(=\) | \(\dfrac{7\times6\times5}{3\times2\times1}\) | |
| \(=\) | \(35\) |
\(35\) ways.
“At least 3 women” in a group of 4 means 3 women or 4 women. Count the first case — 3 women and 1 man — by multiplying the two group choices:
| \(\text{3 women}\) | \(=\) | \({}^{5}C_{3}\) |
| \(=\) | \(10\) | |
| \(\text{1 man}\) | \(=\) | \({}^{3}C_{1}\) |
| \(=\) | \(3\) | |
| \(\text{case 1}\) | \(=\) | \(10\times3\) |
| \(=\) | \(30\) |
Count the second case — 4 women (and no men):
| \(\text{4 women}\) | \(=\) | \({}^{5}C_{4}\) |
| \(=\) | \(5\) |
Add the two cases:
| \(\text{total}\) | \(=\) | \(30+5\) |
| \(=\) | \(35\) |
\(35\) different groups.
Choose from each group separately. First the boys:
| \(\text{boys}\) | \(=\) | \({}^{5}C_{2}\) |
| \(=\) | \(\dfrac{5\times4}{2\times1}\) | |
| \(=\) | \(10\) |
Then the girls:
| \(\text{girls}\) | \(=\) | \({}^{4}C_{2}\) |
| \(=\) | \(\dfrac{4\times3}{2\times1}\) | |
| \(=\) | \(6\) |
The two choices happen together, so multiply:
| \(\text{committees}\) | \(=\) | \(10\times6\) |
| \(=\) | \(60\) |
\(60\) different committees.
Common pitfalls
Frequently asked questions
How do you do combinations when one person must be included?
Fix that person in the selection, then choose the rest: \(r-1\) more members from the \(n-1\) people who remain, which is \({}^{n-1}C_{r-1}\).
How do you do combinations when one person must be excluded?
Remove that person from the pool and choose all \(r\) members from the \(n-1\) who remain, which is \({}^{n-1}C_{r}\).
How do you count “at least” in combinations?
List every allowed case, count each with a product of \({}^{n}C_{r}\) terms, then add the cases. If the unwanted cases are fewer, subtract them from the total instead (the complement).
When do you multiply and when do you add combinations?
Multiply choices made together within one selection, such as choosing some from one group and some from another. Add separate cases, such as the different cases of an "at least" or "at most" condition.
How do you choose a committee with a fixed number from two groups?
Work out the number of ways to choose the required members from each group with \({}^{n}C_{r}\), then multiply the two results.
What does “at most 1” mean in a selection?
"At most 1 of a type" means 0 of that type or exactly 1 of that type. Count both cases and add them.