A shorthand for descending products, and why the empty product is set to one

What to assume they know

• Filling a row of places one at a time from a supply of distinct objects — the descending product and why it has r factors
• Cancelling a common factor from a numerator and a denominator
• Reading an equation in which the unknown sits inside a fraction
• That a definition may be chosen, but is not therefore arbitrary — it can be forced by the need to keep an existing rule true

What they should be able to do

• Write out n! for small n and read the symbol aloud correctly
• State the relation between n! and the factorial one step below it, and use it to rewrite a factorial in cascade
• Say why a peel of two or three factors carries a lower bound on n, and what goes wrong without it
• Evaluate a quotient of factorials by cancelling rather than by computing either one
• Explain why 0! is set to 1, arguing from the relation rather than from convention
• Demonstrate that factorial does not distribute over addition or subtraction, by computing a stated instance
• Solve an equation for an unknown numerator standing over a factorial
• Recognise the general result behind the chapter's two such equations

Where it usually goes wrong

• "3! + 4! = 7!" The chapter puts this question to the student directly, which means it expects the belief. It is false by a wide margin: 30 against 5040. Factorial does not carry across a plus sign, and neither does it carry across a minus sign — Example 5(iii) computes 4920, not 2 and not 2!.
• "0! = 0, because there is nothing to multiply." A product of no factors is taken to be 1 for the same reason a sum of no terms is taken to be 0: it is the value that leaves the neighbouring rules undisturbed. Here the neighbouring rule is 1! = 1 × 0!.
• "0! = 1 is just a convention, so it does not matter." It matters at the next section: the arrangement of all n objects is written as n! divided by 0!, and the answer is n! only because that divisor is 1.
• "You have to work out 12! before you can divide by 10!." You never do. The larger factorial contains the smaller one whole, and cancelling it is the entire technique of this section.
• "(10!)(2!) is the same as 20! or 12!." It is a product of two numbers, 3628800 and 2. The exclamation mark binds to the number it follows, not to the whole expression.
• "n! = n(n − 1)(n − 2)! always." Only for n at least 2. The printed brackets carrying those lower bounds are content, not decoration: you cannot peel off more factors than the number has.
• "7!/5! = (7/5)!" Nothing licenses that. Cancel, do not simplify the argument.

Questions to check understanding

• Evaluate a stated factorial, or a difference of two of them
• True-or-false items on whether factorial distributes over a sum
• Compute a quotient of factorials by cancellation, with a factorial product underneath
• Solve for an unknown standing over a factorial, given a sum of two reciprocals
• Evaluate n!/(n − r)! and n!/(r!(n − r)!) at given n and r, before either has been named
• One-mark items on the value of 0! and on the lower bound needed for a multi-step peel

Examples worth working on the board

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

• The build-up (§6.3.2, p. 105). Four lines printed in a column: 1 is 1!, 1 × 2 is 2!, 1 × 2 × 3 is 3!, 1 × 2 × 3 × 4 is 4!, and so on upward. Show these appearing one at a time; the pattern is the definition.
• The stipulation (p. 105). The chapter states 0! = 1 flatly, as something it is choosing, with no argument attached. That absence is the explanation's opening.
• The cascade (p. 105). Printed across two lines, breaking after the 5 × 4 × 3 × 2! step with the last step set on a second, indented line: 5! rewritten as 5 × 4!, then as 5 × 4 × 3!, then 5 × 4 × 3 × 2!, then 5 × 4 × 3 × 2 × 1!. Every step is the same relation applied once more.
• The peels with their conditions (p. 105). Also printed: n! as n(n − 1)!; next as n(n − 1)(n − 2)!, which the page brackets with a demand that n be at least 2; and finally with a third factor peeled off, so that n, n − 1 and n − 2 all stand outside the residual factorial, bracketed this time with a demand that n be at least 3. Verified: at n = 2 the three-factor peel would call for 2 × 1 × 0 × (−1)!, and the last symbol names nothing this chapter has defined — which is what the bracketed condition is protecting.
• Example 5 (p. 106). Printed: 5! = 120; 7! = 5040; 7! − 5! = 4920. Verified: 5040 − 120 = 4920, and note what it is not — it is not 2!, and not 2.
• Example 6 (p. 106). (i) 7!/5!, printed as 7 × 6 = 42 after the 5! cancels. (ii) 12! over the product of 10! and 2!, printed as 6 × 11 = 66. Verified: 12 × 11 ÷ 2 = 66. Neither factorial is ever evaluated; 12! runs to nine digits and the answer to two.
• Example 7 (p. 106). The expression n! divided by the product of r! and (n − r)!, at n = 5, r = 2. Printed value 10. Verified: 120 ÷ (2 × 6) = 10. Flag this shape for later — it is the combination count, arriving three sections before §6.4 gives it a meaning.
• Example 8 (p. 106). Solve 1/8! + 1/9! = x/10! for x. Printed answer 100. Verified by an independent route: multiplying every term by 10! turns the left side into 10!/8! + 10!/9!, that is 90 + 10, so x = 100 — the chapter clears the denominators the other way round and lands on the same value.
• Exercise 6.2 (pp. 106–107), five items. Q1, evaluate (i) 8! and (ii) 4! − 3!. Q2, asks whether 3! + 4! equals 7!. Q3, compute 8! over the product of 6! and 2!. Q4, solve 1/6! + 1/7! = x/8!. Q5, evaluate n!/(n − r)! at (i) n = 6, r = 2 and (ii) n = 9, r = 5. Verified, arithmetic added here: Q1(i) 40320; Q1(ii) 24 − 6 = 18; Q2 no, since 6 + 24 = 30 while 7! = 5040 — the two sides differ by a factor of 168; Q3 (8 × 7) ÷ 2 = 28; Q4 x = 64, since 8!/6! + 8!/7! = 56 + 8; Q5(i) 6 × 5 = 30; Q5(ii) 9 × 8 × 7 × 6 × 5 = 15120.
• The general result behind Example 8 and Q4. Verified, and mine: an equation of the form 1/(n − 2)! + 1/(n − 1)! = x/n! multiplies up to n(n − 1) + n = n², so x = n² every time. Example 8 has n = 10 and answers 100; Q4 has n = 8 and answers 64. The chapter prints the two problems on facing pages and never remarks that they are one problem. This is the best single item in the topic — it converts drill into a theorem.
• The check that 0! = 1 is forced. Verified, and mine: the relation applied at n = 1 reads 1! = 1 × 0!. Since 1! is 1, the only value 0! can carry is 1. Setting it to 0 would make 1! vanish and would make every quotient with 0! underneath meaningless.

Figures to have open

• A cascade strip for 5!, showing the same number rewritten four ways with the peeled factors accumulating on the left. Standard schematic; the chapter prints this across two lines on p. 105 and it deserves to be shown step by step as one uninterrupted run.
• A cancellation panel for 12! over 10! × 2!, with the shared block of factors shaded in both numerator and denominator. Standard schematic.
• A short table with three columns — n, n!, and n! written as n × (n − 1)! — run from n = 5 down to n = 1 and then to n = 0, so the last row is where 0! = 1 appears. Standard schematic; this table is an added argument, not the book's.
• No textbook figure is required. §6.3.2 carries no numbered figure.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 6 "Permutations and Combinations", §6.3.2 Factorial notation, p. 105. The section is set as a run-in heading rather than on a line of its own.
• Examples 5, 6, 7 and 8, p. 106.
• Exercise 6.2, items 1–5, pp. 106–107.
• Summary, p. 123, which records the factorial product and the relation to the factorial below it, and nothing about 0!.
• Forward pointer inside the same chapter: 0! carries the r = n case of the arrangement formula in §6.3.3, p. 107, and the shape of Example 7 reappears as the combination count in §6.4, p. 116.