Why choices made one after another multiply rather than add

What to assume they know

• Multiplication as repeated addition of equal groups
• Reading a branching diagram: a node, its branches, and the paths from root to tip
• The idea that two descriptions can name the same object, so a list can double-count
• Powers written as repeated multiplication, since the repeats-allowed count becomes one

What they should be able to do

• State the counting principle for two stages and for any finite number of stages
• Justify the product by showing that every outcome of the first stage leaves the same number of second-stage options open, rather than by checking a drawn list
• Count the paths in the chapter's two branching figures and confirm they agree with the product
• Decide, for a given problem, whether stages should be multiplied or cases added, and say what test settles it
• Recompute a count when repetition is permitted, and explain why the factors stop shrinking
• Identify which stage carries a restriction and fill that stage first
• Break a problem with a lower bound on length into disjoint cases, count each by the product rule, and add
• Apply the principle to the chapter's opening lock problem and report the number of sequences to be tried

Where it usually goes wrong

• "Three pants and two shirts make five outfits." Adding counts the garments, not the outfits. The question asks for pairs, and Fig 6.1 shows six of them. Make the student point at the sixth tip.
• "So counting problems are always multiplication." Example 4 adds four numbers. The test is whether the two things being combined are stages of one outcome, which multiply, or alternatives for one outcome, which add — and alternatives may be added only when no outcome belongs to two of them.
• "You can add the four signal counts because they are all signals." You can add them because no signal has two lengths. If the cases could overlap, adding would double-count the overlap.
• "The factors just go down by one each time, always." They shrink only because the supply shrinks. Permit reuse and every factor equals the full supply, which is what turns 24 into 256 in Example 1's Note.
• "Fill the places left to right." Example 3 fills the units place first because that is where the restriction lives, though with reuse permitted either order gives 10. The order starts to matter once repetition is barred: there, filling a restricted place after an unrestricted one can leave a factor that depends on what was already chosen, and the principle needs each stage's count to be a fixed number. Exercise 6.3 Q3 and Q4 are where that bites.
• "The tree proves it." The tree makes the structure visible for six or twelve outcomes; nobody draws 504. The argument that survives the growth is that every first branch has the same number of branches hanging off it — which is true even in Example 1, where the three letters left after the first pick are a different three on every branch.
• "m × n counts unordered pairings." It counts one pant with one shirt as one outcome, and that outcome is a pair of things drawn from two different supplies — a situation where the question of order does not arise yet. It arises in §6.3.

Questions to check understanding

• Count numbers of a given digit-length formable from a stated digit set, with and without reuse
• The same, with a parity or divisibility restriction that forces one place to be filled first
• Count codes, telephone numbers or seat allocations with some places pre-filled
• Count outcomes of a repeated experiment such as tossing a coin a stated number of times
• Count signals or stacks where the number of items used is at least some bound, requiring disjoint cases and a final addition
• One-mark items asking whether two given descriptions are stages or alternatives
• Justify in words why a particular count is a product, referring to equal groups

Examples worth working on the board

Items marked verified are worked out here from the chapter's stated data; the chapter prints no answers to its exercises on these pages.

• The number lock (§6.1, p. 100). Four wheels, each carrying the ten digits 0 through 9. The lock opens on one particular ordered four-digit sequence with no digit used twice. The first digit is known to be 7; the other three are forgotten. Verified: nine digits remain and three ordered places are to be filled, so 9 × 8 × 7 = 504 sequences would have to be tried. The chapter poses this question in its opening paragraph and never returns to it, and no number is given for it anywhere. The 504 is added here.
• Pants and shirts (§6.2, p. 100). Three pants, two shirts, one of each to be worn. Printed count 3 × 2 = 6.
• Fig 6.1 (p. 101). A branching figure: three branches labelled P1, P2, P3, each splitting into two labelled S1, S2, with the six tips written out as P1S1 through P3S2 and the heading above the figure announcing six possibilities. All lettering sits inside the artwork and does not extract. Verified: six tips, and each of the three first branches carries exactly two — this equal-carrying is the visual form of the whole argument.
• The three-stage version (§6.2, p. 101). Two school bags, three tiffin boxes, two water bottles, one of each. Printed working: 2 × 3 = 6 pairs, then 6 × 2 = 12.
• Fig 6.2 (p. 101). The same shape one stage deeper: two branches B1, B2, each splitting into T1, T2, T3, each of those splitting into W1, W2, with all twelve tips written out and the heading announcing twelve possibilities. Verified: I counted the tips on the page image — twelve, six under B1 and six under B2. Also read off the image, since none of this lettering extracts.
• The principle as printed (§6.2, p. 102), stated first for two stages with counts m and n giving m × n, then generalised on the same page to three stages with counts m, n and p giving m × n × p, and stated to hold for any finite number of stages.
• Example 1, ROSE (p. 102). Four distinct letters, four ordered places, no letter reused. Printed working 4 × 3 × 2 × 1 = 24.
• The Note under Example 1 (p. 103). The same four places when letters may be reused. Printed count 4 × 4 × 4 × 4 = 256. Verified: the two counts differ by a factor of 256 ÷ 24, which is not a whole number — 32/3 — so the restriction is not removing some tidy fraction of the arrangements. Show both numbers together.
• Example 2, flags (p. 103). Four flags of distinct colours, a signal being two of them stacked. Printed count 4 × 3 = 12.
• Example 3, two-digit even numbers (p. 103). Digits available: 1, 2, 3, 4, 5, with reuse permitted. Printed working fills the units place first, 2 ways, then the tens place, 5 ways, giving 10. Verified: the even digits among the five are 2 and 4, hence the 2. Because reuse is permitted here, the order is a convenience rather than a necessity: fill the tens place first and it takes 5 digits, after which the units place still takes exactly 2 whatever was chosen, so both factors stay fixed and the answer is 10 either way. Where the order genuinely matters is when repetition is barred — Exercise 6.3 Q3 and Q4, worked in The closed formula, and how allowing repeats changes the count entirely, are the chapter's own such cases.
• Example 4, at least two flags (pp. 103–104). Five flags of distinct colours on a vertical staff; a signal uses two, three, four or five of them in order. Printed counts: 5 × 4 = 20; 5 × 4 × 3 = 60; 5 × 4 × 3 × 2 = 120; 5 × 4 × 3 × 2 × 1 = 120; total 320. Verified: 20 + 60 + 120 + 120 = 320, and the four cases are disjoint because a signal has exactly one length. Note the four- and five-flag counts coincide at 120 — worth pausing on, since the last factor is 1.
• Exercise 6.1 (p. 104), six items, all pure applications of the principle: Q1, three-digit numbers from the digits 1, 2, 3, 4, 5, (i) with reuse and (ii) without. Q2, three-digit even numbers from 1, 2, 3, 4, 5, 6 with reuse. Q3, four-letter codes from the first ten letters of the alphabet, no letter twice. Q4, five-digit telephone numbers from the digits 0 through 9 beginning 6 then 7, no digit twice. Q5, the outcomes recorded when a coin is tossed three times. Q6, signals made from two of five distinctly coloured flags, one above the other. Verified, arithmetic added here: Q1(i) 5 × 5 × 5 = 125 and Q1(ii) 5 × 4 × 3 = 60; Q2, three even digits for the units place then six each for the other two, 3 × 6 × 6 = 108; Q3, 10 × 9 × 8 × 7 = 5040; Q4, two places already fixed and three left to fill from the eight unused digits, 8 × 7 × 6 = 336; Q5, 2 × 2 × 2 = 8; Q6, 5 × 4 = 20.

Figures to have open

• Fig 6.1 (p. 101), redrawn as a schematic: three first-level branches, two second-level branches under each, six labelled tips. The pedagogical point is the identical sub-tree hanging off each first branch, so draw one of them highlighted. Textbook figure; redraw rather than reproduce.
• Fig 6.2 (p. 101), redrawn the same way, three levels deep with twelve tips. Textbook figure. This one is worth working through level by level, because the doubling from six to twelve is the third factor arriving.
• A row of empty boxes for the vacant places, with the shrinking supply written under each box. The chapter draws these boxes inside its worked solutions on pp. 102–105; a clean schematic version is enough.
• A two-panel comparison of 4 × 3 × 2 × 1 against 4 × 4 × 4 × 4 on the same four boxes. Standard schematic.
• No photograph or dataset is needed.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 6 "Permutations and Combinations", §6.1 Introduction, p. 100, and §6.2 Fundamental Principle of Counting, pp. 100–104.
• Fig 6.1 and Fig 6.2, both on p. 101.
• Examples 1 to 4, pp. 102–104, and the Note following Example 1, p. 103.
• Exercise 6.1, p. 104, items 1–6.
• Summary, p. 123, where the two-stage principle is restated.
• Forward pointer inside the same chapter: Example 1 is reused in §6.3 on p. 104 as the first permutation, and again on p. 108 once the notation exists.