# Counting the same result: original mathematical derivation
Author: article_author_oct13
Checked at UTC: 2026-10-13
Scope: All mathematical claims, invented comparisons and three unchanged approved examples in the Japanese/English math-arrangements pair. This is author work, not independent approval. No empirical or external factual claims.

## 1. What is counted
A result is one ordered row when positions matter, one two-person set when order is ignored, or one string when the two occurrences of A are indistinguishable. Two intermediate records may describe the same requested result. If every final result has exactly k intermediate records, the intermediate records partition into disjoint groups of k. Thus intermediate count = k × final count. Division by k requires that equal group size, rather than mere resemblance.

## 2. ma-0281: four distinct books in a row
Label the distinct books P,Q,R,S. Fill four positions without reusing a book. There are 4 choices for position one. For every such choice there are 3 unused books for position two, then 2 for position three and 1 for position four. Each completed row corresponds to exactly one chain of choices, so count = 4 × 3 × 2 × 1 = 24. For a fixed first book, the six tails are the six permutations of the remaining three books (e.g. after P: QRS,QSR,RQS,RSQ,SQR,SRQ). Four different first books give disjoint groups of six.
Factorial notation for a positive integer n is n! = n × (n−1) × ... × 1; hence 4! = 24. Reversing a row changes positions and counts as a different row because all four books are distinct.
All choices: 24 is correct. 16 can be produced by 4×4, allowing reuse in two positions and failing to fill four positions; it does not count these rows. 12 = 24/2 wrongly identifies every row with its reversal; this identification is absent from the prompt. 4 counts only the first-position choice and omits the six tails. These are author-hypothesized errors, not observed user mistakes.

## 3. ma-0282: unordered pair from five people
Label five distinct people P,Q,R,S,T. An ordered pair of different people has 5 × 4 = 20 choices. The two orders (P,Q) and (Q,P) give the same requested unordered pair {P,Q}. Every two-person set has exactly two orders; since its members differ there is no order fixed by swapping. Therefore 20/2 = 10.
Independent enumeration: PQ,PR,PS,PT,QR,QS,QT,RS,RT,ST. Grouping by first member in that fixed alphabetic enumeration gives 4+3+2+1=10.
All choices: 10 is correct. 20 counts ordered pairs, giving every requested result twice. 25=5×5 permits the same person twice and keeps order. 5 counts single people, not all two-person sets.
Invented comparison: if the two people fill distinct roles, chair and recorder, (P,Q) and (Q,P) assign different roles, so there are 20 assignments and division by two is inapplicable. No roles are supplied by the actual exercise.

## 4. ma-0283: strings A,A,B
Temporarily label the two A occurrences A1 and A2. There are 3×2×1=6 labeled rows. Forgetting only the A subscripts yields:
A1 A2 B and A2 A1 B -> AAB
A1 B A2 and A2 B A1 -> ABA
B A1 A2 and B A2 A1 -> BAA.
Every final string has exactly two labeled antecedents. Thus 6/2 = 3 final strings. Independently, B may occupy exactly one of three positions, and once that position is selected both others must be A.
All choices: 3 is correct. 6 retains the artificial distinction between the As. 2 counts the number of letter types, rather than strings. 9=3×3 is a two-choice product with repetition and does not place precisely the supplied three letters. These are possible incorrect operations, not claims about observed behavior.
AAB and BAA are distinct strings; only the subscripts on the As are erased. No quotient by reversal is requested.
Invented comparison: if A1,A2,B are visibly different labeled cards and those labels remain in the result, all six rows remain distinct. This changes the condition and is not a changed bank question.

## 5. Product, grouping and limits
Multiplication here counts consecutive choices whose available counts are 4,3,2,1 (books), or 5,4 (ordered person pairs). It follows by disjoint branching; probabilistic independence is not a prerequisite for this count. Counts of distinct objects need not equal the number of final results.
Division in sections 3 and 4 is justified by groups of exactly two antecedents for EVERY result. Dividing because 'order does not matter' without identifying the groups can be wrong when the identification or group sizes differ. The three examples are straight rows/sets/strings, not circular placements; rotation or reflection identifications are not assumed.
The article counts possibilities only. Turning such a count into a probability needs a probability model and, for a simple favorable-count/total-count ratio, equally likely elementary outcomes; the existing probability guide supplies that distinct task.

## 6. Locale, settings and unchanged inputs
Both languages use the same ordered IDs ma-0281,ma-0282,ma-0283, all revision 1, subject math, topic 確率・統計, level high, difficulty standard; answer IDs and families agree. Wording preserves four distinct books in a row; unordered two-person pairs from five; identical A,A,B strings. No diagram applies. Full canonical prompts/options/explanations and current ledger entries are in bank-excerpts.json. The author has not edited any bank or approval record.
Prerequisite math-counting provides reading of what a quotient counts. Related math-probability and math-chance-models compare outcome definitions and conditional models. They already exist in each locale. The preset includes other matching high/standard probability-statistics questions, not only these three.