Select groups with exact and minimum requirements
Split a restricted selection into disjoint cases or subtract its complement, keeping group membership unordered.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.2. State whether order matters, objects are distinguishable, repetition is permitted and positions or rows are labelled. Count mutually exclusive cases before adding them; account for internal arrangements when using blocks. These lessons include seating in two or more rows. Arrangements in a circle are excluded from this syllabus section. Unless a question says otherwise, people and named roles are distinguishable, and selection is without replacement. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate combinations.
- Use product and sum rules for disjoint choices.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a three-person committee, which B-count cases mean at least one B?
What you will learn
- Count selections with exact numbers from distinct groups.
- Use cases or complements for at-least requirements.
Multiply within a specified composition
From four A-group and three B-group candidates, a three-person committee with exactly two from A and one from B has 4C2 × 3C1 = 6 × 3 = 18 possibilities. The groups are distinct and disjoint, so choosing a pair from A and one person from B specifies a unique committee.
Do not multiply by 3! unless the committee members are given distinct roles. If an individual belongs to both descriptive groups, this simple composition count may double-count or use a person twice; the stated disjoint-group assumption matters. Exactly two from A is one composition, not ‘two or more’.
At least means several disjoint possibilities
For the same seven candidates, at least one B on a three-person committee gives 4C2 × 3C1 + 4C1 × 3C2 + 4C0 × 3C3 = 18 + 12 + 1 = 31. The B-count cases 1, 2 and 3 are disjoint. Equivalently, subtract no B from all committees: 7C3 − 4C3 = 35 − 4 = 31.
A four-person team from five A and four B candidates with at least two of each must contain exactly two A and two B, because its size is four. It has 5C2 × 4C2 = 60 possibilities. Translate every condition together before listing cases; do not add impossible compositions.
Pause and explain
Four A and three B candidates: how many three-person committees have exactly two A?
Worked example
From four A and three B candidates, choose three with at least one B.
Show the worked solution
- All committees: 7C3 = 35.
- Excluded committees contain only A candidates: 4C3 = 4.
- Subtract 35 − 4 = 31; the complement covers every forbidden case exactly once.
Answer 31 committees.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
From four A and three B candidates in disjoint groups, choose three with exactly two A. Count committees.
Give me a hint
The remaining place must be filled by one B candidate.
Compare my reasoning
- Choose two A candidates in 4C2 = 6 ways.
- Choose one B candidate in 3C1 = 3 ways.
- Multiply to obtain 18 unordered committees.
18 committees.
Look for these in your work
- Use the exact composition.
- Do not order committee members.
From four A and three B candidates, choose three with at least one B. Use a complement and check by cases.
Give me a hint
Subtract committees consisting only of A candidates.
Compare my reasoning
- 7C3 − 4C3 = 35 − 4 = 31.
- The disjoint cases with one, two or three B give 18, 12 and 1.
- Their sum is also 31, confirming the complement count.
31 committees.
Look for these in your work
- Include all positive B counts.
- Make the two methods agree.
A four-person team is chosen from five A and four B candidates in disjoint groups. It needs at least two of each. Count teams and explain why no other composition is possible.
Give me a hint
The two minimum counts already add to the team size.
Compare my reasoning
- At least two A and two B fill four places, so the composition is exactly 2 + 2.
- Choose the pairs in 5C2 = 10 and 4C2 = 6 ways.
- Multiply 10 × 6 = 60; any larger count in either group would exceed four places.
60 teams, each with two A and two B.
Look for these in your work
- Resolve the size condition before counting.
- Choose within each disjoint group without order.
Common mistakes
- Keep exactly and at least distinct.
- Use mutually exclusive group-count cases.
- Apply all size conditions before counting.
What can you explain now?
Can a four-person team contain at least two A and at least two B in more than one composition?
Compare with the explanation
No; it must contain exactly two of each.
Their minimum counts already fill all four places.
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Make it stick
Write valid group counts, multiply within each composition, then add cases or subtract the forbidden complement.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.