Cambridge O Level Additional Mathematics · Syllabus 4037 · 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 O Level Additional Mathematics.
Combination in context
A permutation is an ordered count of ways to choose r objects from n distinct objects and place them in r distinct positions, given by \({}^nP_r=\dfrac{n!}{(n-r)!}\); a combination is the same choice without any ordering, given by \({}^nC_r=\dfrac{n!}{r!(n-r)!}\). Which one a question needs is decided by a single test, not by its wording: take the objects chosen and swap two of them. If the swap produces a different outcome, order matters and it is a permutation; if it produces the same outcome, order does not matter and it is a combination. Words such as "select" and "choose" appear in both, so the swap test decides, not the vocabulary.
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 answer that chains a combination into a permutation. Why it fails The syllabus does not require problems that need both methods in one task. Attempting to manufacture such questions in revision teaches a skill that is not assessed, and it blurs the order decision that is. Fix Know the identity \({}^nP_r=r!\,{}^nC_r\) as a relationship and a check, which is required, and keep each individual problem within one method.
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.
Questions students ask about Combination
What is the difference between a permutation and a combination?
A permutation counts ordered arrangements, where swapping two chosen objects gives a different outcome, such as filling named roles like president and secretary; a combination counts unordered selections, where swapping two objects changes nothing, such as choosing an ordinary committee. \({}^nP_r=\dfrac{n!}{(n-r)!}\) counts the first, \({}^nC_r=\dfrac{n!}{r!(n-r)!}\) counts the second. The swap test, not a keyword such as "select" or "choose", decides which formula a question needs.
Why does forgetting the \(r!\) in \({}^nC_r=\dfrac{n!}{r!(n-r)!}\) give the wrong answer?
Without \(r!\) in the denominator, what remains is \(\dfrac{n!}{(n-r)!}\), which is \({}^nP_r\), the permutation, not the combination. The \(r!\) is there because every unordered group of \(r\) objects is counted \(r!\) times among the ordered arrangements, once for each way of arranging that same group. Dropping it does not give a slightly wrong number; it silently answers the ordered question instead of the unordered one, and the answer comes out \(r!\) times too large.
How do you use the identity \({}^nP_r=r!\times{}^nC_r\)?
It says that selecting the group first with \({}^nC_r\) and then arranging its \(r\) members in \(r!\) ways gives the same total as counting ordered arrangements directly with \({}^nP_r\). Use it to check an answer — a permutation should equal \(r!\) times the matching combination — rather than mixing the two formulas together in one calculation. The related identity \({}^nC_r={}^nC_{n-r}\) shortens work whenever \(r\) is more than half of \(n\), since choosing who is in also decides who is left out.
How do you solve an equation such as \({}^nC_2=45\) for \(n\)?
Expand the combination using its cancelled form into a polynomial in \(n\) — here \(\dfrac{n(n-1)}{2}=45\) — and solve that polynomial by ordinary algebra. Because \(n\) counts a number of objects, reject any root that is negative, fractional, or smaller than the lower index of the original expression; only a positive whole number large enough for the combination to exist is a valid answer. Writing that rejection down is part of the solution, not an afterthought.