The closed formula, and how allowing repeats changes the count entirely

What to assume they know

• Filling a row of places one at a time from a supply of distinct objects — the descending product and its r factors
• A shorthand for descending products, and why the empty product is set to one — the factorial relation, cancelling, and 0! = 1
• Solving a quadratic by factorising, and rejecting a root on grounds outside the algebra
• Ratios written as a colon, and turning one into an equation

What they should be able to do

• Reproduce the derivation that turns the descending product into a quotient of factorials
• State the range of r the derivation covers and the range the final statement claims
• Argue why arranging none of the objects should count as one way, and check the formula against that argument
• Evaluate the closed form at r = n and say which earlier definition rescues it
• State the count when objects may be reused, and supply the proof the chapter leaves out
• Compute both counts for the same places and objects, and quantify the gap
• Solve an equation in n built from two arrangement counts
• Solve an equation in r, and discard the root that the counting problem cannot accept
• Handle a leading-digit restriction by subtracting the bad cases from the whole
• Distinguish "all of them together" from its negation, and count the negation as a complement

Where it usually goes wrong

• "The closed form is a better count." It is the same count. Nothing about the number changes; what changes is that the expression now has a denominator you can cancel against another one, which is what makes Examples 12 and 13 tractable.
• "To use n!/(n − r)! you must evaluate both factorials." For n = 52 that is absurd and unnecessary. The denominator always divides out the tail; only r factors survive.
• "Allowing repeats means dividing by something." It means multiplying by more. The count rises, from 24 to 256 on four letters and four places.
• "n to the r and the arrangement count are roughly the same." At r = n they differ by a factor of over ten even for n = 4, and the ratio grows fast with n.
• "Every root of the quadratic is an answer." Example 13 prints two roots and only one of them can be substituted back into the equation it came from. Always return to the range the symbols require: r at most n, and r at least 0.
• "The range 0 ≤ r ≤ n was derived." The upper part was; the lower endpoint was argued on counting grounds and then confirmed against the formula. Presenting it as a consequence of the algebra reverses the chapter's own order.
• "'All the vowels do not occur together' means no two vowels touch." It means it is not the case that all of them are together. Example 14(ii) subtracts only the all-together arrangements, and its answer 36000 counts plenty of strings in which two vowels are side by side.
• "Subtracting bad cases is a trick for hard problems." It is the standard move whenever the forbidden set is easier to count than the permitted one, and Example 11 shows it on a problem with an easy forbidden set: strings that open with 0.

Questions to check understanding

• Evaluate an arrangement count using the closed form, cancelling rather than computing
• Count numbers of a given length from a stated digit set, with and without a leading-zero restriction
• Count arrangements with a parity restriction on the last digit
• Solve for n from a ratio or a multiple relating two arrangement counts
• Solve for r from an equation relating two arrangement counts, stating which roots are admissible
• Count arrangements in which a named group of letters stays together, and the complement of that
• One-mark items on the count when repetition is permitted

Examples worth working on the board

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

• The section's own structure (§6.3.3, p. 107). The heading is followed immediately by the finished formula together with the range 0 ≤ r ≤ n, before any argument is given. The derivation that follows is then stated for r strictly above 0, the r = 0 case is argued separately, and the wider range is restated at the foot of the page. Show this order — the section announces its destination, then earns it in two pieces.
• The derivation (p. 107). The descending product is multiplied above and below by the factors that were left out, running from n − r down to 1. Above, those factors complete the product into n!; below, they are exactly (n − r)!. Verified on n = 6, r = 3: the product 6 × 5 × 4 becomes 6 × 5 × 4 × 3 × 2 × 1 over 3 × 2 × 1, that is 720 ÷ 6 = 120, unchanged.
• The r = n case (p. 107). Printed: the count becomes n! over 0!, hence n!. This is where the previous topic's stipulation does its work.
• The r = 0 case (p. 107). The chapter's argument, in outline: to arrange none of the objects is to leave the whole collection behind, and there is one way to do that; the formula at r = 0 returns n! over n!, which is 1, so the two agree. Note the direction — the count is argued first and the formula is checked against it, not the other way round.
• Theorem 2 (p. 108). With reuse permitted, filling r places from n objects can be done in n to the power r ways. The chapter states the result and says the argument is close enough to Theorem 1's to be left to the reader. Supply it: every place still has the full collection available, because nothing is consumed, so the multiplication principle returns n multiplied by itself r times.
• The three reworked counts (p. 108), each printed once by the formula and once with reuse permitted: four letters into four places, 4! = 24 against 4⁴ = 256; three letters drawn from six, 6!/3! = 120 against 6³ = 216; two posts filled from twelve people, 12!/10! = 132. Verified: 24, 256, 120, 216 and 132 all check. The third has no repeats-allowed partner printed, because one person may not hold both posts.
• The size of the gap. Verified, and mine: at four places from four objects the ratio is 256 ÷ 24, about 10.7; at three places from six it is only 216 ÷ 120, 1.8. The gap widens as r approaches n and closes as r falls, because it is the shrinkage of the supply that is being switched off, and early factors shrink least.
• Example 10 (p. 110): four-digit numbers from the digits 1 to 9, no digit used twice. Printed 9!/5! = 9 × 8 × 7 × 6 = 3024. The solution pauses to say why order matters here, using 1234 and 1324 as two different numbers.
• Example 11 (pp. 110–111): numbers strictly between 100 and 1000 built from 0, 1, 2, 3, 4, 5 with no digit twice. Printed working counts every three-place arrangement, 6P3, then subtracts those beginning with 0, which are 5P2. Verified: 120 − 20 = 100. The page names 092 and 042 as the kind of string being removed.
• Example 12 (p. 111). (i) nP5 = 42 · nP3 with n > 4; the printed working cancels the common factor n(n − 1)(n − 2), which the stated condition on n keeps away from zero, and reaches (n − 3)(n − 4) = 42 and then a quadratic with roots 10 and −3, of which only 10 is kept. Verified: 10P5 = 30240 and 42 × 10P3 = 42 × 720 = 30240 ✓. (ii) nP4 divided by (n − 1)P4 equals 5/3, with n > 4; printed answer n = 10. Verified: 5040 ÷ 3024 = 5/3 ✓.
• Example 13 (p. 112), and the most instructive item in the section: solve 5 · 4Pr = 6 · 5P(r−1). The printed working reaches (6 − r)(5 − r) = 6, then r² − 11r + 24 = 0, then roots 8 and 3, and the printed answer keeps both. Verified at r = 3: 5 × 24 = 120 and 6 × 20 = 120 ✓. An added finding, and it should be taught rather than hidden: r = 8 satisfies the quadratic — (−2)(−3) is indeed 6 — but 4Pr is defined only for r at most 4 and 5P(r−1) only for r − 1 at most 5, so at r = 8 neither side of the original equation names anything. The quadratic is a consequence of the equation, not an equivalent of it, and the admissible range has to be applied afterwards. Treat the printed pair of roots as the chapter's own worked example of the trap.
• Example 14 (p. 112): eight-letter arrangements of DAUGHTER, whose letters are all unlike, with three vowels among them. (i) the three vowels kept adjacent: printed 6! × 3! = 4320, from six objects once the vowel block is treated as one. (ii) the negation of that: printed 8! − 6! × 3! = 36000. Verified: 40320 − 4320 = 36000. The reading of (ii) matters: it is the complement of "all three adjacent", so it includes arrangements in which two vowels sit together and the third does not. It is not the count of arrangements with no two vowels adjacent, which is a different and smaller number.
• Exercise 6.3, items 1 to 9 (p. 114). Q1 three-digit numbers from 1 to 9, no repeats. Q2 four-digit numbers with no repeated digit. Q3 three-digit even numbers from 1, 2, 3, 4, 6, 7, no repeats. Q4 four-digit numbers from 1, 2, 3, 4, 5 with no repeats, and how many are even. Q5 two posts from a body of eight. Q6 find n given (n−1)P3 : nP4 = 1 : 9. Q7 find r given (i) 5Pr = 2 · 6P(r−1) and (ii) 5Pr = 6P(r−1). Q8 all eight letters of EQUATION used once each. Q9 the six letters of MONDAY, (i) four at a time, (ii) all of them, (iii) all of them with a vowel leading. Verified, arithmetic added here: Q1 504; Q2 9 × 9 × 8 × 7 = 4536, the first factor being 9 rather than 10 because a four-digit number cannot open with 0; Q3 3 × 5 × 4 = 60; Q4 120 arrangements, of which 2 × 4 × 3 × 2 = 48 are even; Q5 56; Q6 n = 9, since the ratio collapses to 1/n; Q7(i) the quadratic gives 3 and 10 and only 3 is admissible, with 5P3 = 60 = 2 × 30; Q7(ii) the quadratic gives 4 and 9 and only 4 is admissible, with 5P4 = 120 = 6P3; Q8 8! = 40320; Q9 (i) 360, (ii) 720, (iii) 2 × 5! = 240 since MONDAY has two vowels. Both parts of Q7 throw up an inadmissible root, exactly as Example 13 does — the chapter drills the trap three times without ever naming it.

Figures to have open

• A two-row box diagram: the same r places drawn twice, once with the supply count falling under each box and once with it constant. Standard schematic; this single picture carries sections 7 and 9.
• A cancellation panel for the derivation, with the tail factors shaded identically in numerator and denominator. Standard schematic.
• A number line or bar pair showing 24 against 256 to scale, so the gap is seen rather than read. Standard schematic.
• A strike-through panel for Example 13: the quadratic's two roots, each tested against r ≤ n, one surviving. Standard schematic, and an added framing.
• No textbook figure is required. §6.3.3 and Theorem 2 carry no numbered figure.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 6 "Permutations and Combinations", §6.3.3 Derivation of the formula, p. 107, and Theorem 2 with the reworked counts, p. 108.
• Examples 10 to 14, pp. 110–112.
• Exercise 6.3, items 1 to 9, p. 114.
• Summary, p. 123, which states both counts — the quotient of factorials with the range 0 ≤ r ≤ n, and the power for the repeats-allowed case.
• Backward pointer inside the same chapter: 0! = 1 is fixed in §6.3.2, p. 105, and is what makes the r = n case come out right.