Rewriting the triangle with selection counts, so any row is reachable directly

What to assume they know

• Reading the pattern out of the first few expansions — the coefficient array, the addition rule and the three regularities of §7.2
• Every selection was counted once per arrangement of itself — a selection count, and why it is an arrangement count divided by the arrangements you cannot see
• Choosing what to leave out, and the rule that builds each count from two smaller ones — the rule that builds one selection count out of two smaller ones, and the count of what is left out
• A shorthand for descending products, and why the empty product is set to one — the descending-product shorthand and the convention that fixes its value at zero
• Multiplying out a product of several brackets without collecting like terms

What they should be able to do

• State why building the row for a large index from the triangle alone is impractical, using the chapter's own twelfth-power case
• Write the selection-count formula as §7.2 prints it, including the restriction on the position along the row and the condition on the row number
• State the two end values the page gives, and say what they force at the two slanting sides of the triangle
• Read Fig 7.3 in both directions — from a symbol to its number, and from a number back to the selection it counts
• Explain why the coefficient of a term is the number of ways of choosing which brackets contribute the second letter
• Re-derive the addition rule by splitting the selections into those that use one particular bracket and those that do not
• Explain the left-right symmetry of each row as a restatement of choosing what to leave out
• Write the row for index 7 without writing any earlier row, and use it to expand a seventh power
• Say precisely what this rewriting has established and what still needs proving

Where it usually goes wrong

• "The rewriting is cosmetic — the same numbers in fancier clothes." The numbers are the same and the dependency is not. An entry of the triangle depends on the row above it; a selection count depends on two numbers you already have. That is the whole gain, and it is why row 7 can be written with nothing above it.
• "Any formula would do, as long as it gives the right numbers." The identification has to be justified, and the chapter does not justify it here.
• "You could be asked for a position larger than the row number." The printed range forbids it, and the counting reading says why: there are only so many brackets to choose from.
• "The end entries are 1 by convention." They come out of the formula, once the factorial of zero is taken as 1 — which is the convention Chapter 6 argues for, not this one.
• "The chapter works out the twelfth power." It does not. Checked on every page of the chapter as an image. It names the twelfth power only to price the triangle.
• "Fig 7.3 shows the same rows as Fig 7.1." It carries one more, reaching index 5, and it shows each entry twice over — symbol above value. The doubling is the figure's argument and a redrawn version must keep it.
• "Now the theorem is proved." Nothing here is proved. Two descriptions have been matched against each other for six rows and one counting story has been told. §7.2.1 still has to do the work, and it does it by induction.

Questions to check understanding

• Evaluate a selection count from the formula and place it in the correct row and position of the triangle
• Write out a whole row for a nominated index without building any earlier row
• Expand a bracket to the sixth or seventh power using the row you have written
• Give the coefficient of one nominated term of an expansion, without expanding the rest
• Justify the equality of two selection counts on opposite sides of a row
• Short-answer items asking why the position along a row cannot exceed the row number

Examples worth working on the board

Items marked verified are worked out here from the chapter's printed data; this chapter prints no answers on these pages.

• The motivating target (§7.2, p. 128). The bracket built from 2x and 3y, raised to the twelfth power. The page raises it only to say what the triangle would cost — every row up to index 12 written out first. Checked on all nine page images: this chapter never expands that twelfth power anywhere. It is a cost argument, not a worked example, and an explanation that expands it has answered a question the chapter deliberately leaves open.
• The formula as printed (§7.2, p. 128). The selection count for choosing a given number of objects from a supply, written as a factorial over a product of two factorials, with the position along the row restricted to run from zero up to the row number, and the row number required to be a whole number that is not negative. Both conditions are printed on the same line as the formula and both are content: the position cannot exceed the row number because you cannot choose more brackets than you have.
• The two end values (§7.2, p. 128). The page states outright that the count at position zero and the count at the last position are both 1. Verified from the formula: at position zero the two factorials in the denominator are the factorial of zero and the factorial of the row number, and they cancel the numerator; the last position is the mirror image. This is what puts a 1 down both slanting sides.
• Fig 7.3 (p. 128), captioned with the array's name. Six rows, labelled 0 to 5. Every entry appears twice: the selection symbol on top, its numerical value directly beneath it in brackets. Read off the page image, since the whole figure is artwork. The values are 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; 1 5 10 10 5 1. Verified: Fig 7.3 reaches index 5, one row further than Fig 7.1 and Fig 7.2 on the previous page. That makes the index-5 row the only one the chapter prints both ways — as six bare numbers in the running text of p. 127, and inside a figure here. Rows 0 to 4 appear inside figures and nowhere else.
• Row 7 written straight down (§7.2, p. 128). The page writes the eight selection symbols for row 7 with no earlier row in sight, which is the entire point of the rewriting. Verified: their values are 1, 7, 21, 35, 35, 21, 7, 1.
• The seventh power (§7.2, p. 128). Input: a two-term bracket raised to the seventh, expanded by pairing row 7 with the three regularities of §7.2. The page prints the result with the coefficients left as selection symbols rather than as numbers, which is a deliberate move — it is the last step before the general statement. Verified numerically: the eight coefficients are 1, 7, 21, 35, 35, 21, 7, 1, and the two indices in each of the eight terms add to 7.
• The counting argument (added here, not the chapter's). A bracket raised to the seventh power is seven copies of that bracket multiplied together. Expanding without collecting anything, one term of the product is made by taking one of the two letters out of each of the seven brackets. A term carrying exactly three copies of the second letter therefore arises once for each way of choosing which three of the seven brackets supplied it — and the number of such choices is the selection count. Verified as arithmetic added here on the printed formula, not read off any figure: the entry at position 3 of row 7 works out to 35, which is the count of three-element selections from seven. Be clear with anyone that the chapter prints no numerical value for any entry of row 7 — p. 128 gives that row as eight selection symbols only — and that Fig 7.3 stops at index 5, where the largest printed value is 10.
• The addition rule as counting (added here). Single out one bracket among the eight of an index-8 product. Selections of a given size either use that bracket or they do not; the first kind needs one fewer from the remaining seven, the second needs the full number from the remaining seven. Adding the two counts gives the entry below. Verified against Fig 7.3: 4 and 6 in row 4 stand above 10 in row 5, and 10 is the count of two-element selections from five.
• The symmetry as counting (added here). Naming the brackets that supply the second letter is the same act as naming the brackets that supply the first, so each row must read the same in both directions. Verified against Fig 7.3: every one of the six printed rows does.

Figures to have open

• Fig 7.3 redrawn as a schematic: six rows, index column at the left, each entry showing the selection symbol with its value beneath it. The chapter's own figure (p. 128); the entire figure is artwork and does not extract, so redraw it.
• A bracket-picking movement: n copies of a two-term bracket in a row, one letter lifted from each, the resulting term assembling at the side, and a counter showing how many distinct picks give the same term. Standard schematic, and the central image of this topic — the chapter draws nothing like it.
• A split-by-one-bracket panel for the addition rule: the same row of brackets with one singled out, and the two families of selections drawn separately before being summed. Standard schematic.
• An annotated formula strip carrying the selection count, both of its printed conditions, and an arrow from each condition to the thing it rules out. Standard schematic.
• No photograph or textbook artwork beyond Fig 7.3 is required.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 7 Binomial Theorem, §7.2, p. 128, for the twelfth-power cost argument, the printed selection formula with both of its conditions, the two end values, Fig 7.3 with its caption, the row for index 7 and the seventh power expanded from it.
• §7.2, p. 127, for Fig 7.1 and Fig 7.2, against which Fig 7.3 is read.
• Forward pointer inside the same chapter: the numbered observations after §7.2.1, p. 130, are where this book attaches the name binomial coefficients to these numbers.
• Backward pointer outside this chapter: the selection count itself, the descending-product notation, the rule building one count from two smaller ones and the count of what is left out are all Chapter 6, Permutations and Combinations, §6.4. This chapter recalls them and does not restate them.