Original learning material: a bridge from short-answer counting to proof writing for students exploring France’s Coupe Animath pathway. These are original exercises, not official contest questions or a full mock paper. The methods are useful wherever you study.
You will learn: name a collection precisely, count it in two ways, and explain when an object is counted once, twice or several times. Use this to count pairs, recover missing information and prove that a pattern must exist.
Before you start: multiplication, simple equations, odd and even numbers, and counting without repetition. The combinations lesson and pigeonhole principle help with the later questions. All pair and triple formulas needed here are also explained below.
Choose your pace: allow about 45–60 minutes for the starting checks, theory, worked examples and Levels 1–2. Study the stretch bridge and Level 3 in a second session. Give the proof challenges their own written working time. Understanding why a count is correct matters more than speed.
French vocabulary: double comptage means double counting; paire non ordonnée means an unordered pair. You do not need French to use this lesson.
Three small checks
Try these before reading the solutions. If one is difficult, use its repair and repeat the check with three people or two clubs.
DC-D1
Four pupils A, B, C and D are available. How many different two-pupil teams can be chosen? AB and BA mean the same team.
Solution DC-D1
List AB, AC, AD, BC, BD and CD. There are 6 teams. Alternatively, choose a first pupil in 4 ways and a different second pupil in 3 ways. This gives 12 ordered choices, but each team appears in both orders, so 12 ÷ 2 = 6.
Repair: If you obtained 12, circle AB and BA in your list. They describe one team, not two.
DC-D2
Mina joins two clubs, Nabil joins one and Léa joins three. How many entries of these three pupils’ names appear across all the club lists?
Solution DC-D2
Mina contributes two entries, Nabil one and Léa three. The total is 2 + 1 + 3 = 6. The same person may appear on several lists; we are counting memberships, not different people.
Repair: Write one card for each person–club membership. A person belonging to three clubs needs three cards.
DC-D3
At a gathering, nobody shakes hands with themself and each pair shakes hands at most once. Adding every person’s handshake count gives 18. How many handshakes occurred?
Solution DC-D3
Each handshake is included in the count of each of its two participants. Thus the total 18 counts every handshake twice. There were 18 ÷ 2 = 9 handshakes.
Repair: For just A shaking B, A reports one handshake and B reports one. The sum of the reports is two although the event happened once.
Count the same collection from two directions
Suppose a school wants to count its club memberships. One approach is to ask each student how many clubs they joined and add the answers. Another is to ask each club how many members it has and add those answers. Both methods count the same collection: one record for each student together with a club they belong to.
This is an incidence: a link between two kinds of objects. Here a student is linked to a club by membership. “Incidence” sounds technical, but the object being counted is just a membership card with two labels.
Total memberships counted by students = total memberships counted by clubs.
The totals are equal because every card is counted exactly once on each side. We are not saying that the number of students equals the number of clubs. The first essential step is to name the object: a student, a club, and a membership are three different things.
| Student | Club U | Club V | Club W | Row total |
|---|---|---|---|---|
| A | 1 | 1 | 0 | 2 |
| B | 1 | 0 | 1 | 2 |
| C | 0 | 1 | 1 | 2 |
| D | 1 | 1 | 0 | 2 |
| Column total | 3 | 3 | 2 | 8 |
By students, the total is 2 + 2 + 2 + 2 = 8. By clubs, it is 3 + 3 + 2 = 8. Each 1 lies in exactly one row and one column. This table is a useful picture even when a larger problem gives only totals and no individual memberships.
Pairs: when should we divide by two?
From n different people, choose one person and then a different second person. There are n(n − 1) ordered choices. If the task asks for a two-person team, the orders AB and BA describe the same team. Every team appears exactly twice, so there are n(n − 1)/2 unordered pairs. For n = 1 the formula gives zero, as it should.
Do not divide by two automatically. In a record consisting of a student and a club, the two positions have different roles. Reversing the written labels does not create a second record that your counting method included. Similarly, one game has two players but only one winner. Count the multiplicity for the object you actually chose.
A four-line plan for a counting proof
- Name the object. Write what one record contains, including any order or membership condition.
- Count from the first side. Explain what each person, row, club or pair contributes.
- Count from the second side. Check whether the same record appears once, twice or another fixed number of times.
- Use the comparison. Equate the totals, or compare the required number of records with the available capacity.
Worked example 1: handshakes and two ends
DC-E1. Four people A, B, C and D have handshake counts 3, 2, 2 and 1. Each pair shakes hands at most once, and nobody shakes hands with themself. How many handshakes took place?
Solution DC-E1. Imagine placing one mark at each end of every handshake-line. Counting these marks by people gives 3 + 2 + 2 + 1 = 8. Counting by lines gives two marks per line. Therefore there are 8 ÷ 2 = 4 handshakes. The diagram provides one possible arrangement: AB, AC, BC and AD.
The count attached to a person is often called their degree. A person with no handshakes has degree zero. In any such diagram, the sum of all degrees is twice the number of lines. We do not need every pair to be joined, or every person to have the same degree.
Worked example 2: missing membership information
DC-E2. Twenty students each belong to exactly two of five clubs. Four of the clubs each have seven members. How many members does the fifth club have?
The previous example counted two ends of a line. Here we count a link between a student and a club. Keep those two roles visible before writing an equation.
Solution DC-E2. The twenty students contribute 20 × 2 = 40 memberships. The four known clubs account for 4 × 7 = 28 memberships. The fifth club must account for the remaining 40 − 28 = 12, so it has 12 members.
The answer can exceed the average club size because the clubs need not have equal sizes. It cannot exceed the total number of students; 12 ≤ 20 passes that simple check. We do not divide 40 by two: we want club membership entries, and each such entry has already been counted once.
Worked example 3: count a student together with a pair
DC-E3. Eight students each attend exactly three of six workshops. For every unordered pair of workshops, count their common students. What is the sum of these common-student counts? Prove that some pair of workshops has at least two students in common.
A membership alone has one workshop label. This question mentions a pair of workshops. We therefore enlarge our record: it contains one student and two different workshops that student attends. This change of object is the key step.
Solution DC-E3. A student attending three workshops belongs to 3 × 2 ÷ 2 = 3 different workshop pairs. Each of the eight students therefore contributes three records, giving 8 × 3 = 24 records in total. Counting by workshop pairs instead gives exactly the requested sum of their common-student counts. Thus that sum is 24.
The six workshops form 6 × 5 ÷ 2 = 15 different pairs. If each pair had at most one common student, the sum would be at most 15. Since it is 24, at least one pair has two or more common students. This uses a capacity comparison; it does not claim that every pair has the same number.
To see why several appearances of one student are legitimate, suppose that student attends U, V and W. They contribute once to UV, once to UW and once to VW. These are three different records because their workshop-pair labels differ.
From a count to an impossibility or a maximum
A count can tell us more than a total. If each three-person team uses three pairs and no pair may be reused, then the available pairs limit the number of teams. This is a capacity bound: required records cannot exceed available records.
For a “greatest possible” or “smallest possible” question, a bound is only half the work. You must also produce a valid arrangement reaching it. If your arrangement misses the bound, you have not yet proved that the bound is attainable. In DC-Q8, try listing actual teams after you obtain your upper bound.
A stretch bridge: one object may be counted twice for a less obvious reason
This section prepares DC-Q9 and DC-C3. Treat points and coloured lines as a record of relationships. Lengths, angles and crossings have no role; only which pair is joined and its colour matter.
First count triples
Among n different points, there are n(n − 1)(n − 2) ordered choices of three different points. Each unordered triple occurs in six orders, so the number of triples is n(n − 1)(n − 2)/6. For example, A, B, C can be ordered ABC, ACB, BAC, BCA, CAB or CBA.
Then examine just three points
A mixed corner is a chosen point together with one red neighbour and one blue neighbour. For instance, if AB is red and AC is blue, A is a mixed corner in the triple ABC. Once the centre is fixed, choosing a red neighbour and a blue neighbour gives one corner. Their roles are different, so there is no division by two at this step.
A monochromatic triple has no mixed corners. If a triple is not monochromatic, two of its lines have one colour and the third has the other. At the two ends of that differently coloured line, one incident line is red and one blue: two mixed corners. At the remaining point, its two incident lines have the same colour. Thus each non-monochromatic triple contains exactly two mixed corners.
If one point has r red neighbours and b blue neighbours, it creates r × b mixed corners. Add these products over all points and divide by two to recover the number of non-monochromatic triples. Finally subtract from the total number of triples. Before using this on six points, draw one red-red-blue triple yourself and mark its two mixed corners.
Common counting errors
- The object changes halfway through. Counting students on one side and memberships on the other produces unrelated totals. Write a sample record first.
- Every pair is assumed to exist. The formula n(n − 1)/2 counts all possible pairs. An actual handshake network may use only some of them.
- Dividing by two becomes a habit. Explain exactly which two appearances represent the same object. A membership or a win is not automatically counted twice.
- An average is treated as equality. A total spread over several groups forces a lower bound on at least one group, but does not make all group sizes equal.
- An upper bound is presented as a maximum. Add a construction and check every restriction.
- A drawing supplies an unstated condition. Coloured-line problems here concern pair labels; an intersection without a labelled point has no special meaning.
Practise with feedback
Try six questions. Choose a starting level; the next question adjusts to your answers. Hints and worked feedback are available. Saving progress is optional and stays in this browser.
Interactive practice loads here. You can also use the complete question set below.
Nine graded practice problems
Write your own count before using the choices. DC-Q1–Q3 secure the fundamentals; DC-Q4–Q6 change what is counted; DC-Q7–Q9 require a more deliberate choice of records. Open Hint 1 before Hint 2, and leave the solution closed while you make another attempt. The same nine problems form this lesson’s adaptive practice bank; these visible copies let you work on paper.
Level 1: Build the foundations
DC-Q1
Nine players play one game against each of the other players. Each game has exactly two players. How many games are played?
- 18
- 36
- 72
- 81
Hint 1 for DC-Q1
A game is an unordered pair of different players.
Hint 2 for DC-Q1
Count 9 × 8 player–opponent choices, then ask how many choices describe the same game.
Solution DC-Q1
Answer: 36. Each of the nine players has eight opponents, giving 9 × 8 = 72 player appearances in games. Every game has two players, so it contributes exactly two appearances. There are 72 ÷ 2 = 36 games.
If you got stuck: Do not treat A playing B and B playing A as two games. No player plays themself.
DC-Q2
A grid has four rows and five columns. Its numbers of marked cells in the four rows are 3, 2, 4 and 1. The first four columns contain respectively 2, 3, 1 and 2 marked cells. How many marked cells are in the fifth column?
- 1
- 3
- 4
- 2
Hint 1 for DC-Q2
Count the same marked cells by rows and by columns.
Hint 2 for DC-Q2
The row total is 10. Subtract the marks in the four known columns.
Solution DC-Q2
Answer: 2. Counting by rows gives 3 + 2 + 4 + 1 = 10 marked cells. The first four columns contain 2 + 3 + 1 + 2 = 8 of them. Therefore the fifth column contains 10 − 8 = 2. Each marked cell belongs to exactly one row and exactly one column.
If you got stuck: The two totals count the same cells once each. They should be equal; they should not be added or divided by two.
DC-Q3
Eight people report 3, 3, 3, 3, 2, 2, 1 and 1 handshakes respectively. Nobody shakes hands with themself, and each pair shakes hands at most once. How many handshakes occurred?
- 9
- 8
- 18
- 12
Hint 1 for DC-Q3
Add the reports, but keep track of the two participants in each handshake.
Hint 2 for DC-Q3
The sum is 18, and each actual handshake contributes two to that sum.
Solution DC-Q3
Answer: 9. The reports add to 3 + 3 + 3 + 3 + 2 + 2 + 1 + 1 = 18. Every handshake is counted by both participants, so the number of handshakes is 18 ÷ 2 = 9. The participants need not all have equal counts.
If you got stuck: Use the actual reports. The formula 8 × 7 ÷ 2 applies only if every pair shakes hands.
Level 2: Choose the right object
DC-Q4
Eight people include three siblings. Every pair of people shakes hands exactly once, except that no two of the three siblings shake hands with each other. How many handshakes occur?
- 22
- 24
- 26
- 25
Hint 1 for DC-Q4
Begin with all pairs, then remove precisely the forbidden pairs.
Hint 2 for DC-Q4
There are 8 × 7 ÷ 2 possible pairs. The three siblings form 3 × 2 ÷ 2 pairs among themselves.
Solution DC-Q4
Answer: 25. All eight people form 8 × 7 ÷ 2 = 28 unordered pairs. Among the three siblings there are 3 × 2 ÷ 2 = 3 forbidden pairs. Every other pair shakes hands, so the total is 28 − 3 = 25. A sibling still shakes hands with each of the five other people.
If you got stuck: Subtract pairs among siblings, not every handshake involving a sibling. List the three forbidden pairs if needed.
DC-Q5
Twelve students belong to five clubs. Each club has four members. Every student belongs to either one club or three clubs. How many students belong to three clubs?
- 3
- 4
- 5
- 8
Hint 1 for DC-Q5
Count memberships first, rather than distinct students.
Hint 2 for DC-Q5
Give each of the twelve students one membership. Each three-club student then contributes two extra memberships.
Solution DC-Q5
Answer: 4. The five clubs contain 5 × 4 = 20 memberships. If x students belong to three clubs, the remaining 12 − x belong to one. Thus 3x + (12 − x) = 20, so 12 + 2x = 20 and x = 4. Equivalently, the eight memberships beyond one per student come in groups of two.
If you got stuck: A three-club student contributes two extra memberships after the baseline of one, not three extra memberships.
DC-Q6
Eight players each play every other player once. Every game has one winner and no draw. Five players each win exactly two games. The other three players all win the same number of games. How many games does each of those three players win?
- 4
- 5
- 6
- 7
Hint 1 for DC-Q6
Each game contributes exactly one win to the total, although it has two players.
Hint 2 for DC-Q6
Count all games, remove the ten wins of the five named players, then split the remaining wins equally.
Solution DC-Q6
Answer: 6. There are 8 × 7 ÷ 2 = 28 games and therefore 28 wins. The five players contribute 5 × 2 = 10 wins, leaving 18 for the other three. Each of them wins 18 ÷ 3 = 6 games. These conditions are possible: let each of the three beat all five other players and win once in a three-player cycle. Among the remaining five, arrange the games cyclically so each beats the next two and loses to the previous two.
If you got stuck: A win is attached to just one player. Do not divide the total number of wins by two; that would confuse wins with player appearances.
Level 3: Combine counts and justify bounds
DC-Q7
Nine students each attend exactly four of six different workshops. For each unordered pair of different workshops, record the number of students attending both. Add all fifteen recorded numbers. What is the sum?
- 36
- 54
- 72
- 90
Hint 1 for DC-Q7
A student attending several workshops contributes to several of the recorded intersections.
Hint 2 for DC-Q7
One student attending four workshops contributes once to each of the 4 × 3 ÷ 2 pairs among those workshops.
Solution DC-Q7
Answer: 54. Count records of the form (student, unordered pair of workshops attended by that student). From the workshop-pair side, the requested sum counts every record once. From the student side, each student makes 4 × 3 ÷ 2 = 6 records. Thus the sum is 9 × 6 = 54. A student appearing in several intersections is supposed to be counted several times, once for each different pair.
If you got stuck: Do not count only the 36 memberships. The requested objects contain two workshops and one student, so each student contributes six objects.
DC-Q8
Seven students are available for three-student project teams. Any pair of students is allowed to work together in at most one team. What is the greatest possible number of teams?
- 5
- 6
- 7
- 8
Hint 1 for DC-Q8
Each team uses three unordered pairs of students, and a pair cannot be reused.
Hint 2 for DC-Q8
The seven students provide 21 pairs. An upper bound must be accompanied by a collection of teams attaining it.
Solution DC-Q8
Answer: 7. There are 7 × 6 ÷ 2 = 21 student pairs. Each three-student team uses 3 pairs, and no pair can be reused, so there are at most 21 ÷ 3 = 7 teams. Label the students 1 through 7. The seven teams 124, 235, 346, 457, 156, 267 and 137 attain this bound; for example, 124 means students 1, 2 and 4. Their pair lists are 12/14/24; 23/25/35; 34/36/46; 45/47/57; 15/16/56; 26/27/67; 13/17/37. These are 21 different pairs. Thus exactly 7 teams are possible, and more are impossible.
If you got stuck: A capacity bound proves only “at most”. To answer “greatest possible”, also provide and check a construction with that many teams.
DC-Q9
Six points are given. Each unordered pair is joined by one line, coloured either red or blue. A triple of points is monochromatic if all three lines joining its pairs have the same colour. What is the smallest possible number of monochromatic triples?
- 0
- 1
- 3
- 2
Hint 1 for DC-Q9
Count a mixed corner: one point joined in red to one other point and in blue to a third.
Hint 2 for DC-Q9
If a point has r red neighbours, it has 5 − r blue neighbours and forms r(5 − r) mixed corners. Every non-monochromatic triple contains exactly two mixed corners.
Solution DC-Q9
Answer: 2. There are 6 × 5 × 4 ÷ 6 = 20 triples. At a point with r red neighbours, choose one red and one blue neighbour in r(5 − r) ways. For r = 0,1,2,3,4,5 these values are 0,4,6,6,4,0, so at most 6. Across six points there are at most 36 mixed corners. A non-monochromatic triple has two lines of one colour and one of the other: precisely the two ends of the differently coloured line are mixed corners. A monochromatic triple has none. Thus at most 36 ÷ 2 = 18 triples are non-monochromatic, leaving at least 20 − 18 = 2 monochromatic triples. To attain 2, split the points into two groups of three, colour pairs within a group red and pairs in different groups blue. The two whole groups give red triples. Every other triple uses two points from one group and one from the other, giving one red and two blue lines, so no other triple is monochromatic.
If you got stuck: A triple with mixed colours is counted at two of its points, not three. The minimum also requires a colouring attaining the lower bound.
Three challenges for written reasoning
These are proof tasks. Write complete sentences identifying what is counted and why each count is valid. A numerical answer alone is insufficient. Use the rubrics to review your work; an automatic multiple-choice score cannot judge an arbitrary written proof.
DC-C1: Four corners hidden in a marked grid
A five-by-five grid has exactly three marked cells in each row. Prove that there are two different rows and two different columns whose four intersection cells are all marked. The rectangle need not be a square and its sides need not be next to one another.
Hint 1 for DC-C1
Look at pairs of marked columns within one row, rather than at single marked cells.
Hint 2 for DC-C1
Each row contributes three column pairs. Compare the total with the number of different pairs of columns.
Complete solution DC-C1
Call an occurrence a choice of one row together with two different columns whose cells in that row are marked. Since each row has three marked cells, it contains 3 × 2 ÷ 2 = 3 such column pairs. The five rows therefore give 5 × 3 = 15 occurrences.
There are only 5 × 4 ÷ 2 = 10 different unordered pairs of columns in the entire grid. If every column pair occurred in at most one row, there would be at most 10 occurrences. Since there are 15, some column pair occurs in at least two rows.
Choose those two distinct rows and those two columns. In each chosen row, both chosen columns are marked. Their four intersection cells are therefore all marked, as required. The repeated pair is the same pair of columns in different rows; two unrelated pairs would not give the conclusion.
Self-review rubric DC-C1
- Define the counted object as a row together with a pair of marked columns.
- Count 15 occurrences and only 10 available column pairs.
- Explain why a repeated pair occurs in two different rows.
- Identify all four marked corners; do not assume the rectangle is a square.
DC-C2: When every two clubs share exactly one student
There are v clubs and N students, where v ≥ 2. Every student belongs to exactly two different clubs. Every two different clubs have exactly one student in common. Every club contains exactly k students, where k ≥ 1. Prove that v = k + 1 and N = k(k + 1)/2. Then show that such a system exists for every positive integer k.
Hint 1 for DC-C2
Fix one club. Each of its students names one other club: their second club.
Hint 2 for DC-C2
Explain why that correspondence reaches each other club exactly once. Then count all student–club memberships in two ways.
Complete solution DC-C2
Fix a club A. Every student in A belongs to exactly one other club. No two students in A can name the same second club B, because then A and B would share at least two students, contrary to the condition. On the other hand, every other club B shares one student with A, so every other club is named.
Thus the k students in A correspond exactly to the v − 1 other clubs, one student for each club. Hence k = v − 1, or v = k + 1.
Now count memberships. Counting by clubs gives vk, since there are v clubs with k students each. Counting by students gives 2N, since each of the N students belongs to exactly two clubs. Therefore 2N = vk = k(k + 1), so N = k(k + 1)/2.
For existence, take k + 1 labelled clubs. Create one student for each unordered pair of different clubs, and put that student in precisely the two clubs of their pair. Each student belongs to exactly two clubs. Any two clubs have exactly their pair-student in common. A fixed club pairs with the other k clubs, so it has exactly k students. This constructs the required system for every k ≥ 1. For k = 1, there are two clubs and one student in both, which also satisfies the conditions.
Self-review rubric DC-C2
- Show that a fixed club’s members match the other clubs both without repetition and without omission.
- Deduce v = k + 1.
- Count memberships to obtain 2N = vk.
- Give a construction and check all three conditions, including k = 1.
DC-C3: A seven-point colour challenge
Seven points are given, and every pair is joined by a line coloured red or blue. Prove that at least four triples are monochromatic, meaning that all three pair-lines have the same colour. Construct a colouring with exactly four monochromatic triples. Crossings of lines do not create extra points.
Hint 1 for DC-C3
Count mixed corners as in DC-Q9. At a point with r red neighbours, the number is r(6 − r).
Hint 2 for DC-C3
Each of these numbers is at most 9. The total number of mixed corners is even. For a construction, use the red lines in the diagram below and colour every other pair blue.
Complete solution DC-C3
There are 7 × 6 × 5 ÷ 6 = 35 triples of points. A mixed corner consists of a chosen point, one red neighbour and one blue neighbour. If the point has r red neighbours, then it has 6 − r blue neighbours, so it contributes r(6 − r) = 9 − (r − 3)2 ≤ 9 mixed corners.
Let X be the total number of mixed corners and U the number of non-monochromatic triples. Each non-monochromatic triple has exactly two mixed corners, while each monochromatic triple has none. Thus X = 2U. In particular, X is an even integer.
Seven points contribute at most 7 × 9 = 63 corners. Because X is even, actually X ≤ 62. Hence U = X/2 ≤ 31, and at least 35 − 31 = 4 triples are monochromatic. The parity step matters: a fractional number of triples is impossible.
For a construction, label the points A, B, C, D, E, F, G. Colour the seven cycle lines AB, BC, CD, DE, EF, FG, GA red, and also colour AC, AE, BF and DG red. Colour every other pair blue. A has four red neighbours; every other point has three. Thus the mixed-corner total is 4 × 2 + 6 × (3 × 3) = 62, so this colouring has exactly 35 − 62/2 = 4 monochromatic triples. They are ABC in red and ADF, BEG, CEG in blue. The calculation and this explicit list independently check attainment.
Self-review rubric DC-C3
- Define mixed corners and justify r(6 − r) ≤ 9.
- Explain why every non-monochromatic triple contributes exactly two corners.
- Use the evenness of the total to improve 63 to 62 and deduce four triples.
- Specify every pair’s colour through the red list and blue complement, then verify exactly four monochromatic triples.
Leave with a method, not just a formula
Without looking back, explain why adding everybody’s handshake count counts every handshake twice, but adding everybody’s club count counts each membership once. Then explain why a student attending four workshops contributes to six different workshop-pair intersections.
If the first comparison is unclear, return to the four-person diagram and put a mark at each end of a line. If the second is unclear, write four workshop labels and list their six pairs. If you reached a bound but could not build an example, revisit the full pair list in DC-Q8. Return to a solution-assisted question in a later session and try it without opening the hints.
Next step: study the pigeonhole principle for forced repetition, or counting with bijections for a different way of proving two counts are equal.
France stage context and your next step
The 2026 Coupe Animath first round used a two-hour online integer-answer test. Its autumn notice adopts the spring format, which specifies 25 puzzles. Each correct answer earns one point, with no deduction for a wrong answer. Our short practice sessions teach one method; they are not full timed papers.
The supervised second round lasts three hours for collège and four hours for lycée. The released 2025 paper requires written reasoning for most problems. The final proof challenges here develop that change in style; they do not predict selection-test difficulty. Official tests use French.
Autumn 2026 notice · Spring 2026 format · Official released papers. Format checked 3 October 2026. These original problems are not copied or translated from those papers.
Continue with the extremal principle, return to the combinatorics path, or check the France selection guide.