Group selections & restrictions
Practice Flashcards
Flip to reveal answersHow do you count a group with exactly so many from each category?
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All 8 Flashcards — Group selections & restrictions
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Question
How do you count a group with exactly so many from each category?
Answer
Choose from each group separately, then multiply (AND → ×). e.g. 2 men and 2 women = ⁵C₂ × ⁶C₂.
Question
Why multiply when choosing from two groups?
Answer
Each way of choosing the first group can pair with each way of choosing the second — that's the multiplication principle.
Question
How do you count 'at least 2 women'?
Answer
Add the cases: exactly 2 + exactly 3 + … Each case = choose that many women AND the rest from the other group.
Question
When is the complement (total − unwanted) faster?
Answer
When there are many 'at least' cases but few to exclude — e.g. 'at least 1' = total − (none).
Question
'Exactly 2 men out of a committee of 4' — what about the rest?
Answer
The other 2 must be women, so multiply ⁵C₂ (men) × ⁶C₂ (women) — the numbers add to 4.
Question
'At most 1 girl' on a team of 4 — which cases?
Answer
Exactly 0 girls + exactly 1 girl, added together.
Question
Common slip when choosing from groups?
Answer
Adding the two ⁿCᵣ values instead of multiplying them (it's AND, not OR).
Question
How do you handle 'sum divisible by 3' type group questions?
Answer
Split the numbers into remainder groups (mod 3), then count the choices that make the total work — choosing from each group, multiplying and adding cases.
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Topic 1.10 hub
Counting & binomial (HL only)
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