Arranging given digits to make the largest product

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What to assume they know

Multiplying a two-digit number by a one-digit number
That a two-digit number is ten times its tens digit plus its units digit
Expanding a product over an addition, so that a sum times a number becomes two products added
That the order of two factors does not change their product
Comparing two expressions that share a part, by comparing only the parts that differ
Counting arrangements systematically: three choices for the first slot, then two for what remains

What they should be able to do

Count and list every way three given digits can fill a two-digit-by-one-digit frame, without missing or repeating one
Group the arrangements by which digit is the multiplier, and eliminate one from each group with a reason
Split a two-digit-by-one-digit product into its ten-times part and its one-times part
Explain why the two surviving arrangements have equal ten-times parts, naming the order-of-factors property as the reason
Decide between them by comparing only the one-times parts
Repeat the whole argument with letters for three digits in increasing order, and obtain the general result
State the resulting rule and apply it to a new set of three digits
Say what the argument assumes, and what would happen to it if the digits were not all different

Where it usually goes wrong

"Put the big digits where the big places are, so 53 × 2." This is the instinct the section exists to break. 53 × 2 is 106 and 32 × 5 is 160. The multiplier is a place too, and it multiplies everything, not just one column.
"Make the two-digit number as large as possible." Same error in a different sentence, and it loses by 54.
"The largest digit belongs in the tens place." It belongs outside, as the multiplier — although for the ten-times part it genuinely does not matter which of the two large digits goes where, which is the subtle half of the result.
"You have to check all six." Two structural cuts leave a single comparison, and the second cut needs no arithmetic at all.
"The two survivors differ in their big parts." They do not, and believing they do is what makes students compute rather than compare. The equality is the whole reason the small part gets to decide.
"It only works for 2, 3 and 5." The letter version covers every set of three distinct digits, and the two exercise sets confirm it. This is exactly why the chapter bothers with letters after having already found the answer.
"The smallest digit is wasted." It is placed where its weight is one, which is the best possible use of the digit you can least afford to weight heavily.
"Both parts of the split matter equally." One is ten times the other's scale. That asymmetry is why the comparison collapses to a single small product.

Questions to check understanding

Given three digits, place them in a two-digit-by-one-digit frame for the largest product, and justify the placement without computing every arrangement
Given three digits, place them for the smallest product, and say which step of the argument reverses
List all arrangements of three digits in such a frame and explain why there are exactly that many
Expand a stated two-digit-by-one-digit product into its ten-times and one-times parts
Given two arrangements with equal ten-times parts, decide between them and state the deciding quantity
Prove the general rule for three distinct digits using letters
Explain why swapping the two largest digits between the multiplier and the tens place leaves the ten-times part unchanged

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated inputs. No answer to any exercise the chapter sets the reader is printed anywhere in Part II pp.135–147 — including, notably, the six products of its own opening puzzle — and Part II has no answer-key appendix. (The chapter does print worked answers to its own demonstrations — the 291 decode resolved to 25th December on Part II p.137, the apex-10 and apex-60 pyramids shown completed on Part II pp.138–139, the sum-36 block given as 5, 6, 12, 13 on Part II p.141, and the worked algebra grid resolving to 9 and 5 on Part II p.142 — so do not say the chapter prints no answers at all.)

The puzzle (Part II §6.5, p.142, foot of page). The digits 2, 3 and 5 are to be placed in a frame of two boxes, a multiplication sign, and one box, each digit used once, so as to make the product as large as possible.
The count (Part II §6.5, p.143). The page reasons that the first box can take any of the three digits and each such choice leaves two ways to place the other two, giving six.
The six arrangements, exactly as the page lists them (Part II §6.5, p.143): 23 × 5, 25 × 3, 32 × 5, 35 × 2, 52 × 3, 53 × 2.
The grouping by multiplier (Part II §6.5, p.143): 35 × 2 with 53 × 2; 25 × 3 with 52 × 3; 23 × 5 with 32 × 5.
The three survivors (Part II §6.5, p.143): 53 × 2, 52 × 3, 32 × 5. The page then discards 53 × 2 against 52 × 3 without computing either.
The decisive split (Part II §6.5, p.143). The page writes 32 × 5 as 3 × 10 × 5 together with 2 × 5, and 52 × 3 as 5 × 10 × 3 together with 2 × 3, observes that the first pieces agree, and concludes from the second pieces that 32 × 5 wins.
The six products. Verified, and not printed anywhere in the chapter: 23 × 5 = 115; 25 × 3 = 75; 32 × 5 = 160; 35 × 2 = 70; 52 × 3 = 156; 53 × 2 = 106. The largest is 160. These are worth showing at the very end as a check, never at the start — the chapter deliberately reasons its way there instead, and an explanation that computes all six in the first minute has thrown the topic away.
The general version (Part II §6.5, pp.143–144). Three digits are named with letters in increasing order. The page groups the six products by multiplier exactly as before, reduces to three, eliminates one because the multiplier of the two-digit part is larger in the other, and then splits the last two: one becomes ten times the middle digit times the largest, plus the smallest times the largest; the other becomes ten times the largest times the middle, plus the smallest times the middle. The page notes the first pieces agree and concludes that the winner puts the largest digit outside, as the multiplier, with the remaining pair set inside in decreasing order.
Why the ten-times parts agree, said properly. Verified reasoning, not spelled out in the chapter: both surviving arrangements put the two largest digits in the two slots whose weights multiply to ten — the multiplier and the tens place — so both ten-times parts are ten times the same pair of digits. It is the order-of-factors property alone that makes them equal, and the chapter states the equality without ever saying why it holds. This is section 7, and it is the heart of the topic.
Figure it Out item 1 (Part II §6.5, p.144). The digits 1, 3 and 7 in the same frame. Verified: the six products are 13 × 7 = 91, 17 × 3 = 51, 31 × 7 = 217, 37 × 1 = 37, 71 × 3 = 213, 73 × 1 = 73; the answer is 31 × 7 = 217, which is what the rule predicts.
Figure it Out item 2 (Part II §6.5, p.144). The digits 3, 5 and 9. Verified: the six products are 35 × 9 = 315, 39 × 5 = 195, 53 × 9 = 477, 59 × 3 = 177, 93 × 5 = 465, 95 × 3 = 285; the answer is 53 × 9 = 477. This is the better of the two for teaching, because 95 × 3 and 93 × 5 are the two arrangements a student's instinct reaches for and both lose.
What the argument assumes. Verified remark, mine: the letters are introduced in strict increasing order, so all three digits differ. If two were equal the strict comparisons soften into "not smaller" and the conclusion survives, with more than one arrangement tying for the win. A zero among the digits behaves the same way, and needs the same caveat: decreasing order puts it in the units place, where its weight is one, so the one-times part vanishes — but that is exactly what destroys the final strict step, because the small piece is what the contest was being settled by. With the smallest digit zero, the two leading candidates become equal: for 0, 3, 5 the products are 30 × 5 = 150 and 50 × 3 = 150. So say of zero what the previous sentence says of equal digits — the conclusion survives, but the winner is no longer unique.
The step the argument needs and the Thesis should not assert bare. Verified, mine: going from "the big piece is a product of two digits" to "the two largest digits must fill those two slots" needs one more premise — that no gain in the one-times part can pay for a loss in the ten-times part. For digits p < q < r ≤ 9 it holds, and cheaply: (10qr + pr) − (10pr + qr) = 9r(q − p) > 0, and (10qr + pr) − (10pq + qr) = q(9r − 10p) + pr > 0, since p < q < r forces r ≥ p + 2 and r ≤ 9 together give 9r > 10p. Put this in section 11 — one line — because the explanation teaches the argument, not just its conclusion, and as it stands the asymmetry is only glossed in a misconception bullet.

Figures to have open

The placement tree for section 2. Standard schematic; the chapter argues the count in words and drawing it is what turns "there are six" into something the student has checked.
A proportional split bar for a two-digit-by-one-digit product: a long block for the ten-times part and a short one for the one-times part, drawn to scale. This is the explanation's key figure and is not in the chapter. Drawn to scale it makes the asymmetry of section 8 visible before it is stated.
Two digit tiles that can be swapped between the multiplier slot and the tens slot, with the ten-times value displayed and visibly unchanged. Standard schematic; needed for section 7.
The frame itself — two boxes, a multiplication sign, one box — as the chapter draws it at the foot of Part II p.142. Standard schematic.
No photograph is needed.

Where this sits in the book

NCERT Ganita Prakash Class 8, Part II, printed Chapter 6, "Algebra Play", §6.5 "The Largest Product", Part II pp.142–144. The section opens with the puzzle at the foot of Part II p.142, the numerical argument fills Part II p.143, the letter argument runs from the foot of Part II p.143 to the top of Part II p.144, and the two-item Figure it Out block sits in the middle of Part II p.144.
The Algebra Grids material higher on Part II p.142 is a different topic and is covered by Algebra grids: every row is an equation, and the shapes are the unknowns; §6.6 begins lower on Part II p.144 and is covered by Decoding a divisibility trick with algebra.
The chapter's SUMMARY (Part II p.147) refers to this section as the place where algebra was applied to forming numbers from given digits so as to maximise a product.