Dividing out the swaps you cannot see when some objects are identical

What to assume they know

• A shorthand for descending products, and why the empty product is set to one — factorials, and why r! counts the orderings of r things
• The closed formula, and how allowing repeats changes the count entirely — the closed arrangement formula and the block device for keeping letters adjacent
• Division as splitting a total into equal parts, and what goes wrong when the parts are unequal
• Counting the letters of a word by kind

What they should be able to do

• Explain why the plain factorial overcounts when two objects cannot be told apart
• Label repeated objects temporarily, count, and then quantify the overcount
• Show that every visible arrangement corresponds to the same number of labelled ones, and say why that is what permits the division
• Apply the count for one repeated group, and for several repeated groups at once
• Count arrangements of coloured objects where objects of a colour are alike
• Combine the division with a fixed position, with both ends fixed, and with a group kept adjacent
• Handle a kept-together group that itself contains repeats
• Distinguish the negation of "all together" from "no two together", and count the first as a complement

Where it usually goes wrong

• "Divide by the number of repeated letters." Divide by its factorial. With two alike the two agree and nothing is learned; with three alike, 3! is 6 and the wrong divisor is 3, so the answer comes out doubled.
• "Add up all the repeats and divide by that factorial." ALLAHABAD has six repeated letters in total, but the divisor is 4! × 2!, which is 48, not 6! = 720. Each group is shuffled inside itself; the groups are not shuffled into each other.
• "Subtract the repeats instead." Nothing is being removed from the collection; arrangements are being merged. Merging equal-sized blocks is division.
• "The division works because the numbers happen to come out whole." They come out whole because the blocks are equal, which is the thing that had to be shown. Getting a whole number is a consequence, not evidence.
• "Kept-together groups and repeated letters are two separate tricks." Example 16(ii) needs both in one line, and the vowel block there contains four identical E's, so the block's own internal count is 5! divided by 4!, not 5!.
• "'The four I's do not all come together' means no two I's touch." It is the negation of "all four adjacent" and nothing more. Exercise 6.3 Q10 subtracts only the all-four-together arrangements, and the answer counts plenty of strings with two or three I's side by side.
• "This only applies to words." Example 15 is coloured discs; the same count governs flags, beads, votes and identical machine parts.
• "Labelling changes the problem." It changes the problem temporarily and on purpose, and the division puts it back. The whole method is a round trip.

Questions to check understanding

• Count arrangements of a given word with one or more repeated letters
• Count arrangements of coloured or otherwise identical objects in a row
• Count arrangements with a specified letter fixed at one end, or letters fixed at both ends
• Count arrangements in which a named set of letters stays adjacent, where that set itself contains repeats
• Count the complement of an all-together condition
• Count arrangements with a prescribed gap between two named letters
• One-mark items on what the divisor should be for a stated multiset of letters

Examples worth working on the board

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

• ROOT (§6.3.4, p. 108). Four letters, of which two are the same O. The labelled count is 4! = 24; each visible word answers to 2! labelled ones; the printed result is 24 ÷ 2 = 12.
• The ROOT table (pp. 108–109), which is the section's proof and its best visual asset. Two columns: on the left, labelled strings bracketed in pairs; on the right, the single word each pair collapses to, joined by a long arrow. Verified on the page images: twelve rows in all — ROOT and TOOR on p. 108, then ROTO, TORO, RTOO, TROO, OORT, OROT, OTOR, ORTO, OTRO and OOTR down p. 109 — and every row holds exactly two labelled strings. Twelve rows of two is twenty-four, which is 4!, and twelve is the answer. Show the movement of the arrows firing one row at a time.
• A printed slip in that table, worth knowing before you redraw it. The row that collapses to OORT pairs a first string reading O₁O₂RT with a second reading O₂O₁TR. Strip the subscripts from the second and it spells OOTR, not OORT; its partner should have been O₂O₁RT. Verified on that row. The consequence is that the twenty-four printed strings hold one duplicate and omit one arrangement. The count 12 is untouched and the argument is untouched, but anyone copying the table must fix the row.
• INSTITUTE (pp. 109–110). Nine letters, with I appearing twice and T three times. The labelled count is 9!; each visible arrangement answers to 2! × 3! labelled ones; the printed expression is 9! over 2!3!. The chapter leaves it unevaluated. Verified: 362880 ÷ 12 = 30240.
• Theorem 3 (p. 110). One repeated group of size p among n objects, the rest all unlike: the count is n! divided by p!. Theorem 4 (p. 110), the general case: groups of sizes p₁, p₂, …, p_k, anything left over being unlike, and the count is n! divided by the product of all their factorials. The page states in so many words that both are given without proof. The pictures on the two previous pages are the proof, and section 7 should say so.
• Example 9, ALLAHABAD (p. 110). Nine letters, four A's, two L's, the rest unlike. Printed 9! over 4!2!, evaluated as 7560. Verified: 362880 ÷ 48 = 7560.
• Example 15, discs (pp. 112–113). Four red, three yellow, two green, nine in a row, discs of one colour not being tellable apart. Printed 9! over 4!3!2!, evaluated as 1260. Verified: 362880 ÷ 288 = 1260. Use this to make the point that nothing here is about the alphabet.
• Example 16, INDEPENDENCE (pp. 113–114), four parts, and the richest item in the chapter. Twelve letters with N three times, E four times, D twice. Unrestricted count printed as 12! over 3!4!2! = 1663200. (i) P held at the left end, eleven letters left: printed 11! over 3!2!4! = 138600. (ii) the five vowels — four E's and one I — kept adjacent: the block plus the seven remaining letters make eight objects among which three N's and two D's repeat, giving 8! over 3!2!; inside the block the five vowels rearrange in 5! over 4! ways; printed product 16800. (iii) the negation of (ii): printed 1663200 − 16800 = 1646400. (iv) I pinned to the opening position and P to the closing one, which leaves ten letters between them: printed 10! over 3!2!4! = 12600. Verified, all five: 479001600 ÷ 288 = 1663200; 39916800 ÷ 288 = 138600; (40320 ÷ 12) × (120 ÷ 24) = 3360 × 5 = 16800; the subtraction; and 3628800 ÷ 288 = 12600.
• Exercise 6.3 Q10 (p. 114): the arrangements of MISSISSIPPI in which the four I's do not all stand together. Verified, mine: eleven letters with I four times, S four times and P twice, so 11! over 4!4!2! = 34650 in all; with the four I's fused into one block, eight objects remain among which S repeats four times and P twice, giving 8! over 4!2! = 840; the answer is 34650 − 840 = 33810.
• Exercise 6.3 Q11 (p. 114): arrangements of PERMUTATIONS, whose twelve letters include T twice and nothing else repeated, (i) with P first and S last, (ii) with the vowels all adjacent, (iii) with exactly four letters standing between P and S. Verified, mine: (i) the ten middle letters carry the repeated T, so 10! over 2! = 1814400; (ii) the five vowels are all unlike, and the block plus seven consonants makes eight objects with T twice, so (8! over 2!) × 5! = 20160 × 120 = 2419200; (iii) P and S must sit five places apart, which can happen at seven pairs of positions and in two orders each, fourteen ways, and the remaining ten letters give 10! over 2!, so 14 × 1814400 = 25401600.
• A wrong-divisor demonstration. Verified, and mine: for INSTITUTE, dividing 9! by 2 × 3 instead of by 2! × 3! gives 60480 — exactly twice the right answer, because 3! is twice 3. Put the two side by side; students who divide by the count of repeats rather than by its factorial never notice, since ROOT's only repeat group has 2! = 2 and hides the error.

Figures to have open

• The ROOT collapse table, redrawn: twelve rows, each with two labelled strings on the left, an arrow, and the unlabelled word on the right. This is the chapter's own display on pp. 108–109 and it is the proof, so it must be redrawn rather than summarised — and the OORT row must be corrected to O₂O₁RT before it is used.
• A block-size diagram: n! drawn as a long bar chopped into equal segments, the segment length labelled with the product of the group factorials and the number of segments labelled as the answer. Standard schematic; this is the explanation's picture of why a division is available.
• A wrong-divisor panel for INSTITUTE: 9!/(2!3!) beside 9!/(2 × 3), both evaluated. Standard schematic.
• A vowel-block strip for Example 16(ii), showing the block as one tile among eight and then opening it to show the 5!/4! inside. Standard schematic.
• No numbered textbook figure is required; §6.3.4 has none. The collapse table is unnumbered display matter, not a figure.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 6 "Permutations and Combinations", §6.3.4 Permutations when all the objects are not distinct objects, pp. 108–110, including the ROOT collapse table on pp. 108–109 and Theorems 3 and 4 on p. 110.
• Examples 9 on p. 110, 15 on pp. 112–113, and 16 on pp. 113–114.
• Exercise 6.3, items 10 and 11, p. 114.
• Summary, p. 124, which records the general count with its product of factorials underneath.
• Backward pointer inside the same chapter: the block device used here is set up in Example 14 on p. 112, where the letters are all unlike.