Filling a row of places one at a time from a supply of distinct objects

What to assume they know

• Why choices made one after another multiply rather than add — the multiplication principle and the vacant-places device
• That the count assigned to each stage must be a fixed number, independent of what earlier stages chose
• Substituting particular whole numbers into an expression written with n and r
• Reading a statement of the form "for all n and r satisfying a stated condition"

What they should be able to do

• State what an arrangement in a fixed order is, and say why using only some of the objects still counts as one
• Count orderings of three letters drawn from a six-letter supply directly from the multiplication principle, without a formula
• Reproduce the argument of Theorem 1: r places, and the supply left standing before each is filled
• Explain why the last factor is n − r + 1, and check the claim on particular values of n and r
• Use the factor count as a self-check: r factors, whatever n is
• Evaluate the descending product when r equals n, and say why the final factor is 1
• State the range of r that Theorem 1 as printed covers, and name the value it leaves out
• Read the symbol nPr correctly and say what its two numbers refer to

Where it usually goes wrong

• "nPr means n × r." It means a product of r factors starting at n. For n = 9 and r = 3 that is 504, not 27.
• "The product ends at n − r." It ends at n − r + 1. The reason is countable on fingers: when you arrive at the rth place, r − 1 objects are gone, not r. A student who gets this wrong always produces r + 1 factors, so counting the factors catches it every time.
• "A permutation must use every object." Definition 1 explicitly allows only some of them to be used, and the six-letter word taken three at a time is the chapter's own instance.
• "Theorem 1 covers every r." As printed it requires r to be strictly above 0 and at most n. The upper bound is forced by the supply running out; the lower one is a gap that §6.3.3 later closes on purpose. Do not present the printed range as a slip.
• "The boxes are one method and the P symbol is another." The symbol is a name for the box argument's answer. Every property it has comes from that argument.
• "You must fill the places from left to right." Any fixed order of filling gives the same product, because the supply shrinks by one each time whichever place you are at. What you may not do is let the order depend on your earlier choices.
• "Order counts" means the objects must be arranged alphabetically or by size. It means two different sequences of the same objects are two different outcomes.

Questions to check understanding

• Evaluate nPr for small n and r straight from the descending product
• Count arrangements of a given number of unlike letters or digits taken a stated number at a time
• Fill two named posts from a group, with nobody holding both
• One-mark items on the admissible range of r
• Explain in words why the last factor takes the form it does
• Spot-the-error items presenting a product with the wrong number of factors
• Recognise that a problem stated in words about codes, posts or seats is an ordered-selection count

Examples worth working on the board

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

• What §6.3 renames (p. 104). The four-letter orderings of ROSE counted back in Example 1 are named as orderings of four unlike letters used all at once. The count 24 is carried over unchanged; nothing new is computed, only named.
• Definition 1 (p. 105). An ordering of objects that is fixed and specific, drawn from a collection — all of them, or only some of them. The clause that matters is the second: a permutation need not use everything.
• Three from six (§6.3, p. 104). Three-letter strings, meaningful or not, built from the six unlike letters of NUMBER with no letter used twice. Printed working 6 × 5 × 4 = 120. On the same page, the same three places when letters may repeat: 6 × 6 × 6 = 216.
• Theorem 1 (§6.3.1, p. 105). For r in the range 0 < r ≤ n, and objects that do not repeat, the count is the product starting at n and falling by one at each step, stopping at n − r + 1; this count is written nPr.
• The printed proof (p. 105). A row of r boxes is drawn, annotated as r vacant places. The first takes n objects, the second n − 1, the third n − 2, and the rth takes n − (r − 1).
• The off-by-one, made checkable. Verified: with n = 6 and r = 3 the printed factors are 6, 5, 4, and n − r + 1 = 4 ✓, with three factors. Had the last factor been n − r = 3, the product would run 6 × 5 × 4 × 3 = 360, with four factors — a count three times too large. Put 120 and 360 side by side; the error is not subtle in size, only in appearance.
• The case r = n. Verified: the last factor is n − n + 1 = 1, so the product runs all the way down to 1. For ROSE, n = r = 4 gives 4 × 3 × 2 × 1 = 24, which is Example 1's answer arriving a second time by a different route.
• The five-flag counts as descending products (Example 4, pp. 103–104). Verified: 5P2 = 5 × 4 = 20; 5P3 = 5 × 4 × 3 = 60; 5P4 = 5 × 4 × 3 × 2 = 120; 5P5 = 5 × 4 × 3 × 2 × 1 = 120. The last two agree, and they agree precisely because the fifth factor is 1 — there is only one flag left and only one place for it. This is the cleanest available demonstration that the factor count, not the factor values, is what distinguishes the two.
• Exercise 6.3 Q1 (p. 114): three-digit numbers built from the digits 1 to 9 with no digit used twice. Verified: 9 × 8 × 7 = 504. Worth showing beside §6.1's lock, whose three unknown wheels also draw three ordered digits from nine — the same 504 reached from a problem that looked nothing like it. The chapter does not make this connection; it is added here.
• Exercise 6.3 Q5 (p. 114): from a body of eight people, the posts of chairman and of vice chairman are to be filled, nobody holding both. Verified: 8 × 7 = 56. Two places, so two factors.
• A non-example to show. Ask for 4P6 — six places filled from four unlike objects. Verified: the procedure stalls at the fifth place, where the supply is empty, and this is exactly why Theorem 1 carries r ≤ n rather than leaving r free. The non-example is added here.

Figures to have open

• The row of r vacant boxes with the live supply written under each. The chapter draws this row inside Theorem 1's proof on p. 105 with an arrowed label naming the boxes; redraw it as a schematic and show the supply count.
• A two-panel comparison showing three factors ending at 4 against four factors ending at 3, for n = 6, r = 3, with both products evaluated. Standard schematic.
• A strip showing 5P2, 5P3, 5P4 and 5P5 as four descending products stacked, the last two landing on the same value. Standard schematic; the data are Example 4's.
• No textbook figure is required for this topic. Fig 6.1 and Fig 6.2 belong to the previous topic and Fig 6.3 to §6.4.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 6 "Permutations and Combinations", §6.3 Permutations, p. 104, Definition 1 and §6.3.1 Permutations when all the objects are distinct, p. 105, including Theorem 1 and its proof.
• Backward pointer inside the same chapter: Example 1 on p. 102 and Example 4 on pp. 103–104 supply the arithmetic this section renames.
• Exercise 6.3, items 1 and 5, p. 114.
• Summary, p. 123, which records the permutation count in its closed form only.
• Forward pointer inside the same chapter: the closed form and the r = 0 case are §6.3.3 on p. 107.