Every selection was counted once per arrangement of itself

What to assume they know

• The closed formula, and how allowing repeats changes the count entirely — the closed arrangement count and its range of r
• Dividing out the swaps you cannot see when some objects are identical — dividing a total into equal blocks, and why equality of the blocks is what has to be argued
• A shorthand for descending products, and why the empty product is set to one — factorials and cancelling
• That two descriptions may name one object, so a list can hold the same thing twice under different names

What they should be able to do

• Recognise a question in which reordering the chosen objects does not produce a new answer
• List the selections of two from a small collection and identify the reversals that were deliberately left out
• State the multiplier connecting the arrangement count to the selection count, and say why it is r! and not n!
• Argue that the multiplier is the same for every selection, and explain why that is the step the division depends on
• Derive the closed formula for the selection count from the printed identity
• Evaluate the selection count at r = n and at r = 0, and state the range the final formula claims
• Recognise handshake and chord problems as selections of two
• Say what would go wrong if the objects being selected were not all unlike

Where it usually goes wrong

• "Divide the arrangement count by n!" Divide by r!. Only the chosen objects are being reshuffled; the ones left behind are not being touched. For four objects taken two at a time the divisor is 2, not 24.
• "If order does not matter, the ordered count is the wrong tool." It is the only tool available. The method is to count with order, measure the overcount exactly, and divide it out.
• "The team of X with Y and the team of Y with X are two teams." They are one team with two names. The six-and-six lists on p. 115 exist to make this visible.
• "Selecting nothing has no ways, so the count is 0." Leaving the whole collection alone is something you can do, and there is one way to do it. The chapter fixes the value at 1 for exactly the reason it fixed the arrangement of nothing at 1.
• "A selection count is always smaller than the matching arrangement count." Not at r = 1, where both are n, and not at r = 0, where both are 1. The gap opens only once r! starts exceeding 1.
• "You may divide by r! whenever you want to stop caring about order." Only when every selection really does have r! orderings, which needs the r chosen objects to be unlike one another. If a selection could hold two copies of the same object, its block would be smaller than r! and no single divisor would serve — which is precisely the situation the previous topic had to handle differently.
• "Handshakes and signals are the same kind of counting problem." A signal is an ordered stack and a handshake is an unordered pair. The whole of §6.4 hangs on telling them apart before computing anything.

Questions to check understanding

• Count the teams, committees or groups formable from a stated collection
• Count handshakes among a stated number of people, or chords through a stated number of points on a circle
• Evaluate a selection count from the closed formula, cancelling rather than computing factorials
• Decide, for a question posed in words, whether order matters, and justify it
• One-mark items on the value of a selection count when nothing is selected, or when everything is
• Given an arrangement count and an r, recover the selection count
• Explain in words why the divisor is r! rather than n!

Examples worth working on the board

Items marked verified are worked out here from the chapter's stated data.

• Three players, teams of two (§6.4, p. 114). Players named X, Y and Z; a team takes two of them; the page asks directly whether the team of X with Y differs from the team of Y with X, answers no, and gives 3 as the count.
• Fig 6.3 (p. 115). Three pale ellipses in a row, each carrying one team — XY, YZ and ZX — with the caption beneath. All lettering is inside the artwork and does not extract. Note the caption on this figure is set with a full stop after the abbreviation, unlike the captions of Fig 6.1 and Fig 6.2 on p. 101; cite it as Fig 6.3 either way. Verified: the arrangement count here would be 3 × 2 = 6, and 6 ÷ 2! = 3, which is the figure's three ellipses. The chapter does not perform that division at this point — it simply asserts the 3 — so this is the explanation's chance to make the coming argument feel inevitable.
• Handshakes (p. 115). Twelve people in a room, each shaking hands with every other. The page reasons that X shaking with Y is not a second handshake from Y shaking with X, and identifies the count as selections of two from twelve. Verified: 66. A second route worth showing, and mine: each of the twelve people has eleven handshakes, giving 132 hand-ends, and each handshake owns two of them, so 132 ÷ 2 = 66. The divisor 2 is 2! wearing a disguise.
• Chords (p. 115). Seven points on a circle, joined in pairs. Identified as selections of two from seven. Verified: 21.
• Four objects, two at a time (p. 115). Objects A, B, C, D. The page prints the six selections AB, AC, AD, BC, BD and CD, and then prints the six reversals it has deliberately not listed: BA, CA, DA, CB, DB, DC. Verified: 6 and 6 make 12, which is the arrangement count 4 × 3. This pair of printed lists is the best single asset in the section, because the overcount is written out rather than asserted.
• The r = 2 identity (p. 115). Printed: each selection of two yields 2! orderings, so the arrangement count is the selection count times 2!, and rearranged, the selection count is 4! divided by the product of (4 − 2)! and 2!. Verified: 24 ÷ (2 × 2) = 6.
• Five objects, three at a time (pp. 115–116). The page prints all ten selections and observes that each yields 3! orderings. Set alphabetically, the ten are ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE and CDE. Verified: these are exactly the ten triples drawable from five letters, none missing and none repeated. Note that the page itself prints them in a different order from this one, which is worth saying out loud, because completeness rather than sequence is what a selection list has to achieve. Verified further: 5 × 4 × 3 = 60 and 60 ÷ 6 = 10.
• Theorem 5 (p. 116). For r above 0 and at most n, the arrangement count equals the selection count multiplied by r!. The printed proof is two sentences long: each selection supplies r! orderings, so the selections together supply the selection count times r! orderings, and that total is the arrangement count.
• Remark 1 (p. 116). Rearranging gives the selection count as n! divided by the product of r! and (n − r)!. Also printed there: at r = n the value is 1, since the denominator becomes n! times 0!.
• Remark 2 (p. 116). Selecting nothing at all is defined to happen in one way, by the same argument the chapter used for arranging nothing: to select none is to leave the whole collection behind, and there is one way to do that.
• Remark 3 (p. 116). With that in place the formula also holds at r = 0, so the range printed for the final statement runs from 0 to n inclusive.
• A number that arrived early. Example 7 back on p. 106 evaluated n! over the product of r! and (n − r)! at n = 5, r = 2, and got 10. Verified: that is the count of selections of two from five, computed three sections before the chapter said what it counted. Close the explanation by pointing at it.
• The equality cases, which students get wrong. Verified, and mine: the selection count is smaller than the arrangement count for every r from 2 upward, because r! exceeds 1 there — but at r = 1 both equal n, and at r = 0 both equal 1, because 1! and 0! are both 1. So "always smaller" is false, and the exceptions are exactly the two ends.

Figures to have open

• Fig 6.3 (p. 115), redrawn as three simple enclosures carrying XY, YZ and ZX. Textbook figure; redraw rather than reproduce, and add the discarded reversal beside each enclosure, which the printed figure does not show.
• A collapse diagram for four objects taken two at a time: twelve ordered pairs in a row, arrows merging them in twos onto six selections. Standard schematic; the data are the chapter's two printed lists.
• A fan diagram: one selection of three at the centre, six orderings radiating from it, repeated for two more selections so the equality of the fans is visible. Standard schematic, and this is the picture the argument actually needs.
• A two-column table of r against the arrangement count and the selection count for a fixed n, so the r = 0 and r = 1 rows show them agreeing. Standard schematic; not in the book.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 6 "Permutations and Combinations", §6.4 Combinations, pp. 114–116, up to and including Theorem 5 and Remarks 1 to 3.
• Fig 6.3, p. 115.
• Backward pointer inside the same chapter: Example 7 on p. 106 evaluates the selection count's formula before the section that gives it meaning.
• Summary, p. 124, which records the selection count with the range 0 ≤ r ≤ n.
• Forward pointer inside the same chapter: Remarks 4 and 5 and Theorem 6, pp. 116–117, are the next topic.