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Additional Mathematics · Permutations and combinations

Counting selections with required members

Plain nCr is easy, but 'at least two girls' or 'must include Ali' changes everything.

For selections, the order does not matter, so you use nCr. The extra work is in the conditions: someone must be in, someone must be out, or the group needs a minimum of one type.

This lesson follows counting arrangements with restrictions in the permutations and combinations chapter.

How do I handle a person who must be in or out?

A required person uses up one place, so both n and r fall by one. Choosing 5 from 9 students, with Ali required, means choosing 4 from the remaining 8: 8C4 = 70.

An excluded person only lowers n. Choosing 5 from 9 with Ali out means choosing 5 from 8: 8C5 = 56.

Worked example: at least two girls

A team of 5 is chosen from 5 boys and 4 girls. It must have at least 2 girls. How many teams are possible?

Method 1: case by case. List the allowed numbers of girls and add.

Girls Boys Count
2 3 4C2 × 5C3 = 6 × 10 = 60
3 2 4C3 × 5C2 = 4 × 10 = 40
4 1 4C4 × 5C1 = 1 × 5 = 5

The total is 60 + 40 + 5 = 105.

Method 2: complement. All teams: 9C5 = 126. Teams with fewer than 2 girls: 0 girls gives 1, and 1 girl gives 4 × 5C4 = 4 × 5 = 20. That is 21 unwanted teams.

126 − 21 = 105. The methods agree, which confirms the answer.

The mistake that costs marks

The tempting shortcut is to pick 2 girls first, then any 3 of the remaining 7 people: 4C2 × 7C3 = 6 × 35 = 210. This looks tidy but gives double the true answer.

The reason is that a team with 3 girls is counted 3 times, once for each pair of its girls that could be the “first 2 girls”. The next lesson explains this over-counting in detail.

Check yourself

A committee of 4 is chosen from 3 teachers and 4 students. It must include at least 1 teacher. How many committees are there?

Answer

Use the complement. All committees: 7C4 = 35.

Committees with no teacher: all 4 from students, 4C4 = 1.

The answer is 35 − 1 = 34.

Check by cases: 1 teacher gives 3 × 4C3 = 12, 2 teachers give 3C2 × 4C2 = 18, 3 teachers give 1 × 4C1 = 4. The sum is 12 + 18 + 4 = 34.

What to study next

Continue with distinguishing over-counting from under-counting to see why the shortcut above fails and how to test any count. The word-problem structure worksheet helps you mark the required and excluded members.

If at-least questions are where marks slip, see online one-to-one Additional Mathematics tuition.

Common questions

How do I count a selection that must include a particular person?

Put that person in the group, then choose the remaining members from the rest. If 5 are chosen from 9 and one person is required, choose 4 more from the other 8.

How do I count a selection that must exclude a particular person?

Remove that person from the pool and choose from those left. If 5 are chosen from 9 and one is excluded, choose 5 from 8.

What does 'at least' mean when counting?

It means that number or more. 'At least 2 girls' covers 2, 3, 4 and so on up to the maximum available, so list those cases or use the complement.

When should I use the complement?

Use it when the unwanted cases are fewer than the wanted ones. 'At least 1' is usually the total minus the case with none.

At-least questions offer two valid methods, and a teacher can show you which one suits which question and let you confirm an answer with the other.

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