Choosing what to leave out, and the rule that builds each count from two smaller ones

What to assume they know

• Every selection was counted once per arrangement of itself — the selection count, its closed formula and its range
• A shorthand for descending products, and why the empty product is set to one — cancelling factorials and adding fractions with unlike denominators
• Why choices made one after another multiply rather than add — multiplying across independent stages, and adding across cases that cannot overlap
• Solving a linear equation obtained by equating two reciprocals

What they should be able to do

• State the identity relating a selection count to the count of the complementary size, and justify it by describing a single act two ways
• Use that identity to replace a large lower number by a small one before computing
• State the condition under which two selection counts with the same upper number are equal, and use it to solve for the upper number
• Reproduce the printed proof that one selection count is the sum of two smaller ones
• Give the counting argument for the same identity, based on whether a chosen object is taken
• Split a selection problem into independent stages and multiply, or into non-overlapping cases and add
• Count card selections under restrictions of suit, colour and rank
• Check a case decomposition by confirming that the parts recover the whole

Where it usually goes wrong

• "The complementary identity is a computational trick." It is a statement that two questions are one question. Nothing is being manipulated; the same act is being described from the other end.
• "If two selection counts are equal then the lower numbers must be equal." They may instead add up to the upper number, which is exactly how Example 17 is solved. A student who knows only the first branch cannot do the problem.
• "Theorem 6 says you add the two lower numbers." It says you add the two counts. The lower numbers stay r and r − 1; it is the upper number that goes up by one.
• "The algebraic proof is the explanation." It verifies. The explanation is the question asked of one singled-out object: in, or out. A student who remembers only the algebra cannot rebuild the identity under exam pressure; one who remembers the split can.
• "Cases can always be added." Only when nothing falls in two of them and nothing falls outside them all. Section 12 tests this on Example 19's colour splits, and the test is that the parts recover 270725.
• "Stages can always be multiplied." Only when the count for the later stage does not depend on which choice was made earlier. In Example 18 the women are chosen from the same three whichever man was taken, so the product is safe.
• "'At least one ace' and 'exactly one ace' are the same restriction." Exercise 6.4 Q6 asks for exactly one, which fixes the other four cards outside the aces and makes the count a single product. At-least questions need a sum over cases, and the two answers differ.

Questions to check understanding

• Convert a selection count to its complementary form before evaluating
• Find the upper number given that two selection counts are equal
• Find n from a stated ratio of two selection counts
• Verify or apply the sum identity for particular n and r
• Count committees, teams or selections split by category, multiplying across categories
• Count selections with a fixed number of members drawn from a restricted subgroup, such as bowlers in a team or aces in a hand
• Count selections in which some members are compulsory
• Chord and handshake counts on stated numbers of points or people

Examples worth working on the board

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

• Remark 4 (p. 116, continuing to p. 117). The selection count for the complementary size equals the original one; the page establishes it by writing both as factorial quotients and noticing that the two denominators are the same product in the other order. Immediately afterwards, on p. 117, comes the one-line reading: to pick out r is to throw back n − r. Verified on the chapter's own numbers: Example 18's committee count is 5C3 = 10, and 5C2 is also 10.
• What Remark 4 saves. Verified, and mine: Exercise 6.4 Q1 opens with a selection count whose lower number is 8; converting it to the complementary size turns an eight-factor computation into a two-factor one. This is the practical reason to know the identity, and the chapter never states it.
• Remark 5 (p. 117). If two selection counts with the same upper number are equal, then either their lower numbers are equal, or the two lower numbers add up to the upper one.
• Example 17 (p. 117). Given that the counts at lower numbers 9 and 8 agree, find the count at lower number 17. The printed route equates two factorial quotients and reduces to 1/9 = 1/(n − 8), giving n = 17; then the count at 17 out of 17 is 1. Verified by the shorter route: Remark 5 gives 9 + 8 = 17 at once. Run both — the chapter prints the long one immediately after printing the tool that shortens it.
• Theorem 6 (p. 117). The selection count at lower number r, added to the one at r − 1 with the same upper number, gives the selection count at r from an upper number one larger.
• The printed proof (p. 117), in four moves: write both terms as factorial quotients; pull out the common factor, leaving 1/r in one bracket and 1/(n − r + 1) in the other; add the two fractions, whose numerator becomes n − r + 1 + r, that is n + 1; recognise the result as the required quotient. Verified numerically: 5C3 + 5C2 = 10 + 10 = 20, and 6C3 = 20. And on the card data of Example 19: 13C4 + 13C3 = 715 + 286 = 1001, and 14C4 = 1001.
• The counting argument, which the page does not give. Mine. Single out one object from the n + 1 available. Any selection of r either contains it — and then the other r − 1 come from the remaining n, which happens nC(r−1) ways — or does not, and then all r come from the remaining n, which happens nCr ways. No selection is in both piles and none is outside both, so the two counts add.
• Example 18 (pp. 117–118). Three people to be chosen from a group of two men and three women. Printed: 5C3 = 10 in all. Then, restricted to one man and two women: 2C1 × 3C2 = 6. Verified: 2 × 3 = 6. Note the two operations doing different jobs in one problem — a product because the men and the women are chosen in independent stages.
• Example 19 (pp. 118–119), four cards drawn from a pack of fifty-two, five parts. Printed: the unrestricted count is 52C4 = 270725. (i) all four from one suit, there being four suits of thirteen: 4 × 13C4 = 2860. (ii) one from each suit: the product of four counts of 13C1, printed as 13 to the fourth power and left unevaluated. (iii) all four drawn from the twelve face cards: 12C4 = 495. (iv) two red and two black, there being twenty-six of each: 26C2 squared, that is 325 squared = 105625. (v) all four of one colour: 2 × 26C4 = 29900. Verified: 270725, 2860, 495, 105625 and 29900 all check; and 13 to the fourth power is 28561.
• The decomposition check, and the best item in the topic. Mine, built only from the chapter's own data. Parts (iv) and (v) between them account for the two-two and four-zero colour splits. The two remaining splits are three red with one black and one red with three black, each 26C3 × 26 = 2600 × 26 = 67600. Verified: 105625 + 29900 + 67600 + 67600 = 270725, which is the unrestricted count stated ahead of part (i). A case split that recovers the whole is a case split with no overlap and no gap — the same property Theorem 6 depends on, arriving here as an arithmetic fact the student can check.
• Exercise 6.4 (p. 119), nine items. Q1, given that the counts at lower numbers 8 and 2 agree, find the one at 2. Q2, find n from a ratio of two selection counts, with upper numbers 2n and n and lower number 3 in both, the ratio being (i) 12 to 1 and (ii) 11 to 1. Q3, chords through 21 points on a circle. Q4, a team of three boys and three girls from five boys and four girls. Q5, nine balls chosen from six red, five white and five blue, three of each colour. Q6, five-card selections from a pack containing exactly one ace. Q7, an eleven-player cricket team from seventeen, of whom only five can bowl, with exactly four bowlers required. Q8, two black and three red from a bag holding five black and six red. Q9, a programme of five courses from nine available, two of them compulsory. Verified, arithmetic added here: Q1 the upper number is 10 and the answer is 45; Q2 the ratio reduces to 4(2n − 1)/(n − 2), so (i) n = 5 and (ii) n = 6, which check out as 120 to 10 and 220 to 20; Q3 210; Q4 5C3 × 4C3 = 10 × 4 = 40; Q5 6C3 × 5C3 × 5C3 = 20 × 10 × 10 = 2000; Q6 4 × 48C4 = 4 × 194580 = 778320; Q7 5C4 × 12C7 = 5 × 792 = 3960; Q8 5C2 × 6C3 = 10 × 20 = 200; Q9 the two compulsory courses are fixed, so three more from seven, 7C3 = 35.

Figures to have open

• A keep-or-discard panel: one row of n objects, r of them shaded, with two labels pointing at the same picture — the selection and its complement. Standard schematic; this is the whole of sections 1 to 3 in one image.
• A split diagram for Theorem 6: n + 1 objects with one marked out, and two branches beneath showing selections that include it and selections that do not, each branch carrying its count. Standard schematic, and it is an added argument rather than the book's.
• A small triangle of selection counts for upper numbers up to 6, with one entry shown as the sum of the two above it. Standard schematic; useful, but say plainly that the chapter prints no such array.
• A four-bar stack for the colour splits of Example 19, the four segments totalling the unrestricted count. Standard schematic.
• No numbered textbook figure is required. §6.4 prints only Fig 6.3, which belongs to the previous topic.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 6 "Permutations and Combinations", §6.4 Combinations, Remarks 4 and 5 and Theorem 6 with its proof, pp. 116–117.
• Examples 17 and 18, pp. 117–118, and Example 19, pp. 118–119.
• Exercise 6.4, items 1 to 9, p. 119.
• Summary, p. 124, which records the closed formula for the selection count but neither of this topic's two identities.
• Historical Note, pp. 124–125, which records that a twelfth-century writer knew the complementary identity for particular numbers without holding the general formula.