Permutations and combinations

• Order matters ⇒ \({}^{n}P_{r}\); order does not ⇒ \({}^{n}C_{r}\). “And” multiplies, “or” adds, “not together” is total minus together.
• Order matters ⇒ \({}^{n}P_{r}\), order does not ⇒ \({}^{n}C_{r}\); identical objects divide by \(p!\,q!\); together glue, not together subtract, no two together use gaps.

• 5.2.1 I understand the terms permutation and combination, and solve simple problems involving selections
• 5.2.2 I can solve problems about arrangements of objects in a line, including those involving (a) repetition (e.g. the number of ways of arranging the letters of the word 'NEEDLESS') (b) restriction (e.g. the number of ways several people can stand in a line if two particular people must, or must not, stand next to each other)

Accuracy for this chapter. Every count is an exact integer. Never round a count and never give it to 3 significant figures: \(3360\), not \(3360.0\) or \(3.36 \times 10^3\). Write the method expression before the number — \(\dfrac{8!}{3!\,2!}\), \({}^{6}C_{2} \times {}^{5}C_{2}\), \(4! \times {}^{5}P_{3}\) — so the method is visible even if the arithmetic slips; a bare number from the calculator shows no method. Use the calculator’s \(n\)P\(r\) and \(n\)C\(r\) keys for the arithmetic, but write the expression. Notation: \({}^{n}C_{r}\) or \(\dbinom{n}{r}\) for combinations, \({}^{n}P_{r}\) for permutations. When chapter 25 builds a probability from these counts, that probability is given as an exact fraction or to 3 significant figures; the counts themselves stay exact.

• “A committee of 3 from 10 is \({}^{10}P_{3} = 720\).” Correct Order matters ⇒ permutation; a committee has no order. Swapping two members gives the same committee, so it is \({}^{10}C_{3} = 120\). Posts (chair, secretary, treasurer) do have an order: \({}^{10}P_{3} = 720\).
• “A and B not together: take them out and arrange the other six.” Correct Not together \(=\) total \(-\) together. For eight people: \(8! - 2 \times 7! = 40\,320 - 10\,080 = 30\,240\).
• “NEEDLESS has \(8!\) arrangements”, or “\(\dfrac{8!}{3! + 2!}\)”. Correct Identical letters are counted repeatedly; divide by the product \(3! \times 2! = 12\): \(3360\).
• “A and B together: glue them, \(7!\).” Correct The unit can be AB or BA: \(2 \times 7!\). Multiply by the arrangements inside the unit, unless the glued objects are identical, when there are none.
• “At least 2 women: choose 2 women, then any 2 of the other 9.” Correct That counts a committee with 3 women several times over. Split into the cases 2W, 3W, 4W, count each, and add.
• “No two girls together \(=\) total \(-\) all three girls together.” Correct That is not all together, which still allows two girls side by side. No two together needs the gaps method.
• “A committee of 3 from 10 is \({}^{10}P_{3} = 720\).” Repair A committee has no order: \({}^{10}C_{3} = 120\).
• “Chair, secretary and treasurer from 10 is \({}^{10}C_{3}\).” Repair The posts are different, so order matters: \({}^{10}P_{3} = 720\).
• “NEEDLESS has \(8! = 40\,320\) arrangements.” Repair The three Es and the two Ss are identical; divide by \(3!\,2!\): \(3360\).
• “NEEDLESS \(= \dfrac{8!}{3! + 2!}\).” Repair Divide by the product \(3! \times 2! = 12\), not the sum: each visible arrangement is counted \(6 \times 2\) times.
• “A and B together: \(7! = 5040\).” Repair The glued unit can be AB or BA; multiply by \(2!\): \(10\,080\).
• “A and B not together: \(6!\).” Repair Not together \(=\) total \(-\) together \(= 8! - 2 \times 7! = 30\,240\).
• “The three Es together in NEEDLESS: \(\dfrac{6!}{2!} \times 3!\).” Repair The Es are identical, so there is nothing to arrange inside the unit: \(\dfrac{6!}{2!} = 360\).
• “At least 2 women from 6 women and 5 men, committee of 4: \({}^{6}C_{2} \times {}^{9}C_{2}\).” Repair Choosing 2 women and then any 2 of the other 9 counts some committees more than once (that gives 540). Split into 2W, 3W, 4W and add: \(265\).
• “2 women and 2 men: \({}^{6}C_{2} + {}^{5}C_{2} = 25\).” Repair The two choices are made together (“and”), so they multiply: \(15 \times 10 = 150\). Only separate cases (“or”) add.
• “No two girls together \(=\) total \(-\) all three girls together.” Repair That is “not all together” (4320), which allows two girls side by side. Use the gaps method: \(1440\).
• “Two rows with the two tallest at the back: \(8! \times {}^{6}C_{2}\).” Repair Select who is in each row, then arrange each row: \({}^{6}C_{2} \times 4! \times 4! = 8640\). The selection is not an extra factor on \(8!\); the answer must be smaller than \(8!\).
• “\({}^{n}C_{2} = 45\) gives \(n = 10\) or \(-9\).” Repair \(n\) counts objects; reject \(-9\) and say why: \(n = 10\).
• “\(0! = 0\).” Repair \(0! = 1\), so that \({}^{n}C_{n} = \dfrac{n!}{n!\,0!} = 1\): there is one way to choose everything.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 25 turns these counts into probabilities: when you reach it, re-answer worked example 2 as “the probability that a random committee has at least two women”. Chapter 26 uses \({}^{n}C_{r}\) as the binomial coefficient: re-read the combination definition there. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Chapter 24 · Probability & Statistics 1 · How it is assessed
• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Probability & Statistics 1 content, examined in Paper 5. Paper 5 (Probability & Statistics 1) is 40% of an AS Level that includes it and 20% of the A Level, for which it is compulsory. Its questions use no algebraic methods beyond the Paper 1 content, and it is the foundation for Paper 6. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A question sets one scenario (a word, a committee, a row of people, a set of digits) and asks for the total, then for counts under added restrictions, sometimes as a “show that” with the count given. The reasoning is in the choice of move: which objects to glue, which places to fill first, which cases to add.
• MF19 gives \(\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}\), with the binomial series in the Pure Mathematics section. \(n!\), \({}^{n}P_{r}\), the rule for identical objects and every restriction technique must be known; see the MF19 card.
• Every count is an exact integer: never rounded, never to 3 significant figures. Write the expression, such as \(\dfrac{8!}{3!\,2!}\) or \({}^{6}C_{2} \times 4! \times 4!\), before the value; an unsupported calculator answer earns nothing. A probability built from counts (chapter 25) is an exact fraction or 3 significant figures.