Decide what makes two results different
Read a row of books from left to right, recording which book occupies each position. With four distinct books, reversing the row gives a different arrangement. When forming a pair of people, the completed pair does not retain who was chosen first. Read whether positions or roles remain in the result, as well as whether the question says arrange or choose.
In a string containing two identical A's, you may attach temporary labels to the A occurrences to make counting easier. Those labels do not appear in the finished string. Ask whether a distinction introduced for your calculation belongs in the requested answer.
Fill the positions with the books still available
Any of the four books can occupy the first position. Once it is placed, three books remain for the next position, then two, then one. Every completed chain of choices gives a different row, so there are 4 × 3 × 2 × 1 = 24 arrangements. Fixing the first book leaves six orders for the remaining three; each of the four first-book choices has those six continuations.
The product is written 4!, read as four factorial. For a positive integer n, n! is the product of the integers from n down to 1. Here multiplication counts successive choices of positions. It does not require an assumption of probabilistic independence.
Dividing 24 by two gives 12 if a row and its reversal are treated as identical. The exercise specifies a row and supplies no such identification. Counting only the first-position choice gives four, leaving out the six possible continuations for each choice.
Group the two selection orders for each pair
Label the five people P, Q, R, S and T. Choosing one person and then a different person gives 5 × 4 = 20 ordered records. Choosing P then Q and choosing Q then P both finish as the same pair PQ. Every pair has exactly two selection orders, so there are 20/2 = 10 unordered pairs.
A list checks the result: PQ, PR, PS, PT, QR, QS, QT, RS, RT and ST. There are four pairs containing P, three containing Q but not P, then two and one. This nonrepeating enumeration gives 4 + 3 + 2 + 1 = 10.
Under a different invented condition, assigning the two people to the roles of chair and recorder makes P-as-chair with Q-as-recorder different from the reverse assignment. There would be 20 assignments. The actual exercise assigns no roles and asks for unordered pairs, so its division by two is justified.
Erase the temporary labels on the identical A's
Temporarily call the two A occurrences A1 and A2. With these labels retained, the three objects have 3 × 2 × 1 = 6 rows. Erasing the labels maps both A1A2B and A2A1B to AAB. Likewise A1BA2 and A2BA1 become ABA, while BA1A2 and BA2A1 become BAA.
Every finished string has two labeled rows corresponding to it, giving 6/2 = 3 strings. You can also count the three possible positions for B: first, middle or last. Once B's position is chosen, both remaining positions must be A.
Only the temporary A labels are erased. AAB and BAA remain distinct strings because their letter positions differ. If visibly different cards A1, A2 and B retained their labels in the result, all six rows would remain distinct. That is an invented comparison with a changed condition, rather than the bank exercise.
Explain the divisor through records of each result
An unordered pair had two selection orders; a string with the identical A's had two assignments of temporary labels. Division by two works because every requested result groups exactly two intermediate records. Do not divide merely because some objects look alike: explain how many times each final result was counted.
Changing the result definition need not preserve the same divisor. Treating reversed rows or rotated circular placements as identical introduces conditions absent from these exercises. Preserve the distinctions that the stated result retains.
We are counting possible results here. Converting a count to a probability also needs a trial model and information about how likely its elementary outcomes are. Knowing there are ten pairs does not by itself establish that a real selection process chooses all ten with equal probability.
Revisit the exercises as records and final results
The exercises below proceed from distinct books to unordered pairs to a repeated-letter string. Before opening an answer, define one result, then count choices for successive positions or write a short nonrepeating list. After reading, hide the explanation and reconstruct what multiplication counted and which two records were grouped when dividing by two.
If you are unsure what a quotient or remainder counts, return to the prerequisite “Counting objects: groups, gaps and endpoints.” Continue with the related “Probability: list the outcomes before counting” to examine equal-likelihood conditions. “Probability conditions: does the next draw change?” compares ordering and drawing rules. The practice preset selects high-school-level, standard probability and statistics questions, including matching questions beyond these three examples.