Cambridge IGCSE Additional Mathematics · Syllabus 0606 · Permutations and Combinations
Permutation
What is Permutation?
A permutation of r objects taken from n distinct objects is an ordered selection: r of the objects are chosen and then placed in r distinguishable positions, so that changing the order produces a different outcome. The number of such permutations is written n P r and equals n factorial divided by (n minus r) factorial, which is the product of the r largest factors of n factorial. It is defined for non-negative integers with r no greater than n, no object may be used twice, and it is the count to use whenever the positions carry distinct roles, ranks or places.
This definition is part of the Permutations and Combinations chapter in Cambridge IGCSE Additional Mathematics.
Common mistakes with Permutation
- Choosing the formula from a keyword instead of from the order decision. Why it fails “Select”, “choose” and “pick” appear in both kinds of question. Select three students to be president, secretary and treasurer is a permutation despite the word “select”. Vocabulary describes the context; only the swap test describes the structure. Fix Name two of the chosen objects, swap them, and ask whether the result is a different outcome. Write the one-sentence conclusion before writing anything else.
- 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.
- Mis-writing the permutation as \(\dfrac{n!}{r!}\) or \(\dfrac{(n-r)!}{n!}\). Why it fails \(\dfrac{n!}{r!}\) removes the wrong tail: it leaves \(n-r\) factors instead of \(r\), so \({}^8P_3\) would come out as \(8\times7\times6\times5\times4=6720\). The inverted form gives a fraction below \(1\), which cannot be a count at all. Fix Count the factors. \({}^nP_r\) must have exactly \(r\) of them. Draw the slots if in doubt: \(r\) slots, \(r\) numbers.
- Practising circular arrangements as if they were required content. Why it fails Circular problems count rotations of the same seating as one outcome, which changes \(n!\) into \((n-1)!\) and needs an argument this syllabus does not ask for. Fix The required version seats people in a row, where the places are genuinely distinguishable and the count is a plain permutation.
- 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 Permutation
- Write the classification down. A student who writes “order matters here because the three prizes are different, so this is a permutation” and then miscounts is in a far better position than one who writes a bare number that happens to be wrong. In a topic where the whole solution is often two lines long, the temptation to do it all on the calculator is at its strongest — and on Paper 2, where the keys exist, it is the most expensive habit here.
- 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.