Cambridge IGCSE Additional Mathematics · Syllabus 0606 · Permutations and Combinations
Combination
What is Combination?
A combination of r objects taken from n distinct objects is an unordered selection: r of the objects are chosen and nothing distinguishes one chosen object from another, so swapping two of them leaves the outcome unchanged. The number of such combinations is written n C r, or as the binomial coefficient n choose r, and equals n factorial divided by the product of r factorial and (n minus r) factorial. The r factorial in the denominator is present because each unordered group of r objects appears r factorial times among the ordered arrangements, so the arrangement count must be divided by it. It is defined for non-negative integers with r no greater than n.
This definition is part of the Permutations and Combinations chapter in Cambridge IGCSE Additional Mathematics.
Common mistakes with Combination
- Forgetting the \(r!\) in the denominator of a combination. Why it fails What is left is \(\dfrac{n!}{(n-r)!}\), which is the permutation. So a “combination” with a dropped \(r!\) does not produce a slightly wrong number — it silently answers a different question, and produces an answer exactly \(r!\) times too big. Fix Remember where \(r!\) comes from: every unordered group of \(r\) was counted \(r!\) times among the arrangements, so it must be divided out. If you know the reason, you cannot lose the symbol.
- Using \({}^nC_r\) when the positions are named or distinct. Why it fails \({}^nC_r\) throws away exactly the information the question depends on. President-Ama and President-Ben are different outcomes; a combination cannot tell them apart, so the answer is \(r!\) times too small. Fix Look for anything that distinguishes one chosen object from another — a title, a rank, a numbered seat, a place in a code. If there is one, use \({}^nP_r\).
- Choosing the formula from a keyword: “it says select, so it must be a combination”. Why it fails “Select”, “choose” and “pick” describe how the objects are taken, not what happens to them afterwards. Select three students to be president, secretary and treasurer is an arrangement. Fix Read to the end of the sentence, then swap two chosen objects and ask whether the outcome changed. The verb never decides; the structure does.
- Building a single count by chaining a combination into a permutation — choose the committee, then give its members distinct roles. Why it fails Outcome 11.3 does not require a counting task whose one answer needs both methods chained. Manufacturing such questions in revision teaches a skill that is not assessed, and it blurs the order decision that is. Fix Keep each individual count within one method. The identity \({}^nP_r=r!\,{}^nC_r\) is still required and still useful — converting between two counts that a question asks for separately, as a hence part does, is not the same thing as chaining them to build one.
Examiner tips on Combination
- What to notice while you play with it. Fix \(n=10\) and step \(r\) from \(0\) to \(10\). The combination values rise to a peak at \(r=5\) and then fall symmetrically — that is \({}^nC_r={}^nC_{n-r}\) made visible. The permutation values, by contrast, never come back down: they climb steeply and then flatten, because each extra slot multiplies the count by the number of objects still available — and at the last slot that number is \(1\), so \({}^{10}P_9\) and \({}^{10}P_{10}\) are equal. If you can predict that shape before you see it, Section E has done its job.
- What all three have in common. Each opened with a single sentence settling the order decision for the whole question, and each later part reused an earlier number instead of starting again. Those two habits — decide once, reuse always — save more time and prevent more errors than any amount of extra speed at evaluating combinations.