For a selection with a condition, split into cases, calculate each case as a product of combinations, then add the cases. Because only the group matters, every case uses nCr. This is the second main skill in permutations and combinations, after deciding whether order matters.
How do you split a condition into cases?
Read the condition word by word.
- Exactly k of a type: one case. Choose k from that type and the rest from the other type.
- At least k of a type: a list of cases: k, k + 1 and so on, up to the most possible.
- At most k of a type: a list of cases: 0, 1, up to k.
- Must include a person: place the person, then choose the rest from everyone else.
- Must not include a person: remove the person from the pool first.
Within one case, separate selections are multiplied. Across different cases, you add.
Worked example
A group of 5 is chosen from 6 boys and 4 girls. Find the number of groups with (a) exactly 2 girls, and (b) at least 2 girls.
Part (a): exactly 2 girls means 2 girls and 3 boys.
4C2 × 6C3 = 6 × 20 = 120
Part (b): at least 2 girls means 2, 3 or 4 girls. The group size is 5, so the number of boys is 3, 2 or 1.
| Girls | Boys | Working | Groups |
|---|---|---|---|
| 2 | 3 | 4C2 × 6C3 = 6 × 20 | 120 |
| 3 | 2 | 4C3 × 6C2 = 4 × 15 | 60 |
| 4 | 1 | 4C4 × 6C1 = 1 × 6 | 6 |
Total = 120 + 60 + 6 = 186
Check (complement): all groups of 5 from 10 is 10C5 = 252. Groups with fewer than 2 girls: 0 girls gives 6C5 = 6, and 1 girl gives 4 × 6C4 = 4 × 15 = 60. So 252 − 6 − 60 = 186. The answers agree.
The mistake to watch for
A common slip is to add when the two selections happen together.
Mistaken answer to (a): 4C2 + 6C3 = 6 + 20 = 26
The student treats “choose girls” and “choose boys” as alternatives.
The group needs both the girls and the boys. For every one of the 6 ways of choosing the girls, there are 20 ways of choosing the boys, so the counts multiply. The correction is 6 × 20 = 120.
Check yourself
1. A committee of 4 is chosen from 7 men and 5 women. How many committees have exactly 2 women?
Show answer
2 women and 2 men: 5C2 × 7C2 = 10 × 21 = 210.
2. A team of 3 is chosen from 8 students, and Hana must be in the team. How many teams are there?
Show answer
Hana is placed, then 2 more are chosen from the other 7: 7C2 = 21.
3. A group of 4 is chosen from 6 boys and 3 girls. How many groups have at least 3 boys?
Show answer
3 boys and 1 girl: 6C3 × 3C1 = 20 × 3 = 60. 4 boys: 6C4 = 15. Total = 60 + 15 = 75.
Check: all groups 9C4 = 126. Fewer than 3 boys: 2 boys and 2 girls gives 15 × 3 = 45, and 1 boy and 3 girls gives 6 × 1 = 6. So 126 − 45 − 6 = 75.
Where this leads next
When a question has a long list of cases, it is quicker to subtract: see using a complement to simplify counting. Then use the mixed practice set. The non-calculator working trainer helps with evaluating nCr by cancelling.
Students sometimes list the cases correctly in class but miss one under time pressure. A teacher in online one-to-one Additional Mathematics tuition can watch how you set the cases out and help you build a habit that does not miss them.