Permutations and Combinations

• Swap two chosen objects. Different outcome means arrangement and \({}^nP_r\); same outcome means selection and \({}^nC_r\). Everything else in this chapter is arithmetic, and the arithmetic is factorial cancellation.
• Every error in this chapter is silent. Nothing will tell you that you counted arrangements when the question wanted groups — only the habit of running the swap test first, naming the operation before using it, and comparing the answer against the unrestricted count.
• Swap two chosen objects. Different outcome means arrangement; same outcome means selection. The numbers \(n\) and \(r\) cannot tell you which — only the question can, and only if you read it to the end.
• \(n!\) counts the orderings of \(n\) objects, \(0!=1\) because the empty ordering is one ordering, and \(n!=n\times(n-1)!\) is the only manipulation you need. A quotient of factorials is a short product — never a big number divided by another big number.
• Draw \(r\) slots. Write \(n\), then \(n-1\), then \(n-2\), one per slot, and multiply. That product is \({}^nP_r=\dfrac{n!}{(n-r)!}\), it has exactly \(r\) factors, and it assumes no object is ever reused.
• \({}^nC_r\) is \({}^nP_r\) with the orderings divided out: \(\dfrac{{}^nP_r}{r!}=\dfrac{n!}{r!\,(n-r)!}\). The \(r!\) is not decoration — it is the number of times each group was counted, and remembering that is why you will never write the formula without it.
• \({}^nP_r=r!\,{}^nC_r\) — select, then arrange. \({}^nC_r={}^nC_{n-r}\) — choosing who is in chooses who is out. Use the first to check every answer and the second whenever \(r\) is more than half of \(n\).
• Satisfy the restriction first, then permute what is left, then multiply. Check the answer two ways: it must be smaller than the unrestricted count, and the restricted count plus its complement must return that count exactly.

• Decide whether order matters by swapping two chosen objects and asking whether the outcome changed, and justify the decision in words before writing a formula. Section A
• Use factorial notation, evaluate \(n!\) for small \(n\), state and justify \(0!=1\), and cancel a factorial quotient such as \(\dfrac{8!}{5!}\) without expanding it. Section B
• Build and use \({}^nP_r=\dfrac{n!}{(n-r)!}\) from the slot model, and know it counts ordered selections of \(r\) objects from \(n\) distinct objects with no repetition. Section C
• Build and use \({}^nC_r=\dfrac{n!}{r!\,(n-r)!}\), and explain the \(r!\) in the denominator as the number of times each unordered group was counted. Section D
• Use the two identities \({}^nP_r=r!\,{}^nC_r\) and \({}^nC_r={}^nC_{n-r}\) to check an answer and to shorten a calculation. Section E
• Solve arrangement problems in an everyday context, including those in which a particular object must be included, must be excluded, or must occupy a particular position. Section F
• Solve selection problems, including committees drawn from two groups with a stated composition. Section G
• Handle restrictions by multiplying independent choices, adding mutually exclusive cases, or subtracting the unwanted count from the total, and say which of the three you are using. Section H
• Solve an algebraic counting equation such as \({}^nC_2=45\) by expanding to a quadratic in \(n\) and rejecting any root that is negative or not a whole number. Section I
• State the syllabus boundary — repeated objects, circular arrangements and any single count that chains a combination into a permutation — and recognise a question that lies outside it. Syllabus boundary
• Recognise and repair the standard errors of this topic, above all choosing a formula from a keyword rather than from the order decision. Mistake clinic

Notice the two end rows. The case with \(0\) students uses \({}^5C_0=1\) and the case with \(4\) students uses \({}^6C_0=1\). Those are not decorative: they are the reason the decomposition works arithmetically, and they come directly from \(0!=1\) in Section B. A student who believes \({}^nC_0=0\) will find the two end cases vanish and the total fall short by \(20\).

• 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.
• Writing \(0!=0\). Why it fails \(0!\) counts the arrangements of an empty collection, and there is exactly one — the empty arrangement. Setting it to \(0\) puts a zero in the denominator of both end cases: \({}^nP_n=\dfrac{n!}{0!}\) and \({}^nC_n=\dfrac{n!}{n!\,0!}\) would each be a division by zero, so “arrange all \(n\) objects” and “choose all \(n\) objects” would both stop having an answer at all. Fix \(0!=1\). Test it: \({}^5C_5\) must be \(1\), and \(\dfrac{5!}{5!\,0!}=1\) only if \(0!=1\).
• 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.
• Adding two counts that overlap, or multiplying two cases that should be added. Why it fails Adding requires the cases to be mutually exclusive, so that no outcome is counted twice. Multiplying requires the choices to be consecutive and independent, so that every outcome is built exactly once. Using the wrong operation turns a valid decomposition into a number with no meaning. Fix Say the sentence out loud. “First this and then that” multiplies. “Either this or that, never both” adds. If you cannot say either sentence cleanly, the split is wrong.
• Counting a restriction directly and then subtracting its complement as well. Why it fails Complementary counting is an alternative route: total minus unwanted. Doing both and combining them removes the wanted outcomes twice. It usually shows up as an answer close to zero, or negative — which at least is visible. Fix Pick one route and finish in it. Use the other afterwards as an independent check that must give the same number.
• Using \({}^nP_r\) for a team, committee or group with no roles. Why it fails \({}^nP_r\) counts the chosen objects together with an ordering of them. A team has no ordering, so every team is counted once for each of its \(r!\) internal orderings, and the answer is \(r!\) times too large. Fix Run the swap test. If swapping two chosen members leaves the same team, divide the arrangement count by \(r!\) — that is, use \({}^nC_r\).
• 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.
• Counting the same group once for each order it could be written in. Why it fails This is the previous error seen from the inside. Listing \(\{A,B,C\}\), \(\{A,C,B\}\), \(\{B,A,C\}\) and so on as separate outcomes over-counts by \(r!\), because a set has no first element. Fix Adopt a convention when listing by hand — always write the chosen objects in alphabetical or numerical order — and each group then appears exactly once.
• Writing \(0!=0\). Why it fails \(n!\) counts arrangements, and there is exactly one arrangement of nothing — the empty one. Setting \(0!=0\) puts a zero in the denominator of both end cases: \({}^nP_n=\dfrac{n!}{0!}\) and \({}^nC_n=\dfrac{n!}{n!\,0!}\) both become a division by zero, so neither “arrange all \(n\) objects” nor “choose all \(n\) objects” has an answer any more. Fix \(0!=1\). The recursion forces it: \(1!=1\times0!\), and \(1!=1\).
• Writing \(n!=n(n-1)\) and stopping. Why it fails The product runs all the way down to \(1\). \(5!\) is \(120\), not \(20\). What has actually been written is \({}^nP_2\), so the error silently substitutes one counting quantity for another. Fix Say the definition aloud with its ending: “\(n\) times \(n-1\) times \(n-2\), all the way down to one”.
• Forgetting the \(r!\) in the denominator of \({}^nC_r\). Why it fails What remains is \(\dfrac{n!}{(n-r)!}\), which is \({}^nP_r\). The answer is not slightly wrong; it is the answer to the other question, \(r!\) times too big. Fix Remember why it is there: each group was counted \(r!\) times among the arrangements. Then test on \({}^4C_2\), where the six pairs can be listed by hand.
• 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.
• Evaluating \(12!\) and \(9!\) separately in order to find \(\dfrac{12!}{9!}\). Why it fails It is not wrong, but it is slow, error-prone and often impossible: \(20!\) exceeds what many calculators will display exactly, and a rounded intermediate value can corrupt an answer that must be an exact integer. Fix Cancel first. \(\dfrac{12!}{9!}=12\times11\times10=1320\), with no large number anywhere.
• Evaluating \({}^nP_r\) or \({}^nC_r\) with \(r>n\). Why it fails \((n-r)!\) would be the factorial of a negative integer, which does not exist. It is also plainly impossible: you cannot fill \(5\) slots from \(3\) objects without reusing one, and reuse is forbidden. Fix If a case in a decomposition needs \(r>n\), that case simply cannot occur and contributes \(0\) — drop it, do not force it.
• Accepting a negative or fractional value of \(n\) from a quadratic. Why it fails \(n\) is the number of objects available. There is no set with \(-9\) members and none with \(7.5\). The root satisfies the algebra and not the situation, and only the situation is being asked about. Fix Write the rejection explicitly: “\(n=-9\) is rejected because \(n\) cannot be negative”. The rejection is part of the answer; one that lists both roots has not finished.
• Accepting a positive integer root without checking it is large enough. Why it fails If the equation contains \({}^nC_4\), a root of \(n=3\) is invalid even though it is a positive integer, because \({}^3C_4\) does not exist. Fix State the minimum permissible \(n\) beside the expansion, before solving, so the test is already written down when the roots appear.
• Adding two counts that describe successive stages: \({}^5C_2+{}^6C_2=25\) for “two students and two adults”. Why it fails Adding counts the ways of doing one or the other. The committee needs both, so each pair of students must be paired with each pair of adults, which is a product. Fix Say the sentence: “first choose the students and then choose the adults”. “And then” multiplies. The answer is \(10\times15=150\).
• Multiplying two counts that are alternatives: \(60\times5\) for “exactly \(3\) girls or exactly \(4\) girls”. Why it fails Multiplying builds one outcome out of two decisions. Here the two cases are different kinds of outcome and no team is in both, so nothing is being built — the counts should be pooled. Fix “Either this or that, never both” adds. The answer is \(60+5=65\).
• Adding cases that overlap, such as “at least \(2\) girls” plus “at least \(2\) boys”. Why it fails A team with \(2\) girls and \(2\) boys satisfies both descriptions, so it is counted twice. The excess is invisible — the total is simply too large by the size of the overlap, and looks like an ordinary number. Fix Define cases by an exact value (exactly \(0,1,2,3,4\) girls). Every outcome then lands in exactly one case, and the full set must sum to the unrestricted total.
• Counting a restriction directly and subtracting its complement, then combining the two. Why it fails They are two routes to the same number, not two contributions. Adding them doubles the answer; subtracting one from the other gives zero. Fix Answer with one route. Use the other silently as a check: the two must agree, and if they do not, one of them is wrong.
• Designating particular objects as “the guaranteed ones”: pick \(1\) boy, pick \(1\) girl, then pick any \(2\) more. Why it fails The finished team does not record which boy was the guaranteed one, so the same team is produced several times over by different choices. For \(4\) from \(6\) boys and \(5\) girls this gives \(6\times5\times{}^9C_2=1080\), against a total of only \({}^{11}C_4=330\) teams in existence. Fix Split by exact composition, or take the complement — both give \(310\). And always compare the answer with the unrestricted count.
• Allowing a symbol or digit to repeat when the question forbids it — using \(9^4\) where \({}^9P_4\) was wanted. Why it fails \(n^r\) keeps the full pool available at every slot; \({}^nP_r\) removes each object once used. For \(n=9\), \(r=4\) they differ by more than a factor of two (\(6561\) against \(3024\)), and both look like reasonable answers. Fix Look for the phrase “no digit repeated” or “all different”. If your reasoning needs \(n^r\), re-read the question — repetition problems are outside the 0606 requirement.
• Practising repeated-object arrangements, such as the letters of LEVEL, as if they were required content. Why it fails Nothing about the mathematics is wrong, but it is not examined at 0606, and the correction factor it needs is not part of the syllabus. Every hour on it is an hour not spent on outcomes 11.1 to 11.3. Fix Recognise it and move on. See the syllabus boundary; the required version of the same context uses a word whose letters are all different.
• 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.

• 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.
• If you meet one anyway. Older textbooks and general-purpose worksheets are not written to this syllabus, so an excluded problem will occasionally appear in practice material. Do not conclude that your revision is incomplete. Check the exercise against this page: if it turns on repeated identical objects, on a circle, or on chaining a selection into an arrangement, it is outside the requirement, and skipping it costs you nothing.
• 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.
• The one-line reminder to carry forward. Write it on the inside cover of your notes: swap two, then decide. Those four words prevent the one error that no amount of careful arithmetic can repair, and they require remembering no formula at all.

• Cambridge IGCSE Additional Mathematics 0606 is assessed by two compulsory written papers. Both last two hours, both carry 80 marks and half the qualification, both mix structured and unstructured questions, and both may set questions on any part of the subject content — so a counting question can appear in either, and every candidate sits both. The papers differ in one way that matters a great deal here: Paper 1 is a non-calculator paper, and a scientific calculator is required for Paper 2. Plan for both. On Paper 2 the \(n!\), \({}^nP_r\) and \({}^nC_r\) keys are available — and that convenience is its own trap, because the calculator will evaluate whichever of the two you press and has no opinion about which one the question wanted. On Paper 1 there is no key at all, and the same answers have to be produced by cancelling factorials on paper.
• A counting question arrives as a short paragraph of context and a number to find. The work is in three layers, and a complete answer contains all three:
• Layer 1 · Classify Decide, in writing, whether order matters, and state \(n\) and \(r\). One line, which computes nothing and determines everything after it.
• Layer 2 · Structure Say what is being multiplied or added and why: “choose the students, then choose the adults”, or “case 1 plus case 2”. This sentence is the method.
• Layer 3 · Evaluate Substitute, cancel, and give a single whole number. No units, no rounding, no decimal point.
• 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.