Venn Diagrams, Distributions, Routes & Scheduling

Roughly half the DILR paper is neither a pure table nor a pure seating puzzle: it is a Venn diagram of survey respondents, a grid of who bought how many of what, a network of roads or pipes, or a timetable with precedence rules. These sets are worth 3 marks a question with −1 for a wrong MCQ, and they reward a candidate who recognises the family in the first thirty seconds — because each family has one correct opening move, and after that move the set is bookkeeping. A Venn set is solved the moment you have labelled its seven regions; a distribution set the moment you have drawn the grid with its row and column totals; a route set by adding at every node; a schedule by finding the longest chain of dependencies. This chapter gives each opening as numbered steps you can rehearse, and then works one complete set per family with every line shown, so that the method is something you have watched run rather than something you have read about.

Venn Diagrams — Seven Regions and Four Formulas

Three sets A, B, C divide the population into seven regions inside the circles and one outside. Label them by what they contain: three "only" regions (only A, only B, only C), three "exactly two" regions (A and B but not C, and so on), and one centre (all three). Every number the set gives you is a sum of some of these regions, and every question asks for a sum of some others. Inclusion–exclusion connects them: |A ∪ B| = |A| + |B| − |A ∩ B| for two sets, and |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |C ∩ A| + |A ∩ B ∩ C| for three. The trap in the three-set formula is that "|A ∩ B|" means everyone in both A and B, including those also in C — it is two regions, not one.

The four counts a Venn question asks for (S₂ = sum of the three pairwise intersections, T = all three)
CountFormulaRegions it adds up
At least one (the union)|A| + |B| + |C| − S₂ + TAll seven inside regions
Exactly one|A| + |B| + |C| − 2·S₂ + 3·TThe three "only" regions
Exactly twoS₂ − 3·TThe three two-way regions
At least twoS₂ − 2·TThe three two-way regions plus the centre
  1. Draw the three circles and write the centre first. The "all three" number is the only region given directly; everything else is derived from it.
  2. Fill the three two-way regions by subtracting the centre from each pairwise intersection: (A and B only) = |A ∩ B| − T.
  3. Fill each "only" region by subtracting its three neighbours from the set total: only A = |A| − (A and B only) − (A and C only) − T.
  4. Add the seven regions and compare with the population to find "none". If the sum exceeds the population, a number has been double-counted — go back to step 2.

Worked set. Of 120 students, 60 study Physics, 50 Chemistry and 70 Mathematics; 25 study Physics and Chemistry, 30 Chemistry and Mathematics, 35 Physics and Mathematics; 15 study all three. Centre T = 15. Two-way regions: P and C only = 25 − 15 = 10; C and M only = 30 − 15 = 15; P and M only = 35 − 15 = 20. Only regions: only P = 60 − 10 − 20 − 15 = 15; only C = 50 − 10 − 15 − 15 = 10; only M = 70 − 15 − 20 − 15 = 20. Sum of the seven = 15 + 10 + 20 + 10 + 15 + 20 + 15 = 105, so 15 study none. Check by inclusion–exclusion: 60 + 50 + 70 − 25 − 30 − 35 + 15 = 105 ✓. Exactly two = S₂ − 3T = 90 − 45 = 45; exactly one = 180 − 2 × 90 + 3 × 15 = 45; at least two = 90 − 30 = 60. Consistency: 45 + 45 + 15 = 105 ✓.

⚠️ "Both" does not mean "only both"
When a set says "25 study Physics and Chemistry", that is |P ∩ C| and includes the 15 who also study Mathematics; the P-and-C-only region is 10. When a set says "25 study Physics and Chemistry but not Mathematics", that is the region itself and the centre must be added to get |P ∩ C| = 40. Read every intersection statement for which of the two it is before writing a single number — the entire set is wrong if you guess.

Maximum and Minimum Overlap

When the intersections are not given, the set asks for their extremes. For two sets in a population of N: maximum |A ∩ B| = the smaller of |A| and |B| (put the smaller set entirely inside the larger), and minimum |A ∩ B| = |A| + |B| − N, or 0 if that is negative (spread the two sets out as far as they will go). Worked: in a class of 100, 70 like tea and 80 like coffee. Maximum both = 70. Minimum both = 70 + 80 − 100 = 50: the 30 who dislike tea and the 20 who dislike coffee are at most 50 different people, so at least 50 like both.

For three sets the same "count the people who miss something" argument gives minimum |A ∩ B ∩ C| = |A| + |B| + |C| − 2N (or 0). Worked: of 80 students, 60 passed Maths, 50 passed Science and 70 passed English. Those who failed something: 20 + 30 + 10 = 60 failures at most, and each failure removes at most one person from the all-three group, so at least 80 − 60 = 20 passed all three. By the formula: 60 + 50 + 70 − 160 = 20 ✓. The maximum who passed all three is the smallest set, 50. Between these two bounds every value is achievable, so a question asking "which of the following cannot be the number who passed all three?" is answered by the option outside 20–50.

  • Percentages behave exactly like counts with N = 100. "70% like tea, 65% like coffee, 10% like neither" gives union = 90%, so both = 70 + 65 − 90 = 45%. The wrong answer 35% uses N = 100 and forgets the 10% who like neither.
  • To maximise "exactly one", minimise every overlap; to maximise "none", maximise every overlap. The two extremes are the same diagram read from opposite ends, so a set usually asks both — solve them together.
  • Construct the extreme case; do not just quote the formula. A TITA answer of 50 is safe only if you can place 50 people in the centre and still honour every total. The construction is the proof, and it also catches a mis-copied number.

Distribution Grids — Counting Under Constraints

A distribution set has people down the side, items across the top, and a number in each cell — how many of each item each person has. Unlike a matching puzzle, the cells are quantities, so the grid carries row totals and column totals, and those totals are the strongest conditions in the set. The method is the same every time: draw the grid with its totals, enter every cell that a condition fixes outright, then use each total to fix the last unknown in its row or column. When two cells remain unknown in a line, a condition elsewhere will decide one of them.

  1. Draw the grid with a totals row and a totals column. Write the grand total in the corner and check that the row totals and the column totals both add to it — if the set gives both, this check is free.
  2. Enter the zeros first. "P bought no mangoes" fixes a cell at 0 and is the most useful kind of condition, because it removes a whole unknown from two totals at once.
  3. Convert relations into equations on the row. "R bought twice as many apples as bananas" with a row total of 6 and no mangoes means 2x + x = 6, so x = 2: bananas 2, apples 4.
  4. Use each column total to close the last unknown in it. Then re-read the conditions you have not yet used — one of them is usually what breaks the remaining tie.

Worked set. Four friends P, Q, R and S each bought exactly 6 fruits from apples, bananas and mangoes. In total 10 apples, 8 bananas and 6 mangoes were bought. P bought no mangoes. Q bought equal numbers of each fruit. R bought twice as many apples as bananas and no mangoes. S bought at least one of each fruit. Solution. Q = (2, 2, 2). R: mangoes 0, so apples + bananas = 6 with apples = 2 × bananas, giving R = (4, 2, 0). Mango column: 6 = 0 + 2 + 0 + S, so S bought 4 mangoes and therefore 2 other fruits; "at least one of each" forces S = (1, 1, 4). Apple column: 10 = P + 2 + 4 + 1, so P bought 3 apples; P's row of 6 with no mangoes gives 3 bananas: P = (3, 3, 0). Banana check: 3 + 2 + 2 + 1 = 8 ✓. Grid complete and unique.

The completed distribution grid (apples, bananas, mangoes)
FriendApplesBananasMangoesTotal
P3306
Q2226
R4206
S1146
Total108624
🎯 Why the last condition was needed
Without "S bought at least one of each", S could be (2, 0, 4) or (0, 2, 4) and P would shift to match — two valid grids, and a question on P's bananas would have no unique answer. When a set feels solved with a condition unused, that is the signal to look for a second grid; the unused condition is either the tie-breaker or a check. CAT sets are built so that every condition earns its place.

Routes, Networks and Points Tables

Counting shortest routes in a grid. A shortest route from the bottom-left corner to a point m blocks east and n blocks north uses exactly m eastward and n northward steps in some order, so the number of routes is the number of ways to choose which n of the m + n steps go north: C(m + n, n). For 4 east and 3 north: C(7, 3) = 35. When the grid has a missing road or a blocked junction, the formula fails but the addition method does not: write 1 on every junction along the two starting edges, and at every other junction write the sum of the numbers immediately west and immediately south of it. The number at the destination is the count. A blocked junction simply gets 0.

Worked set. From A, the destination B is 4 blocks east and 3 blocks north; junction P is 2 east and 1 north of A. Routes through P = (routes A → P) × (routes P → B) = C(3, 1) × C(4, 2) = 3 × 6 = 18. Routes avoiding P = 35 − 18 = 17. Both numbers appear as options for whichever the question asked, so read whether it says "through" or "avoiding". Network capacities. When pipes or roads carry a maximum flow each, the most that can pass from source to sink is limited by the tightest cut: any set of pipes whose removal disconnects source from sink. Worked: S → A carries 8, S → B carries 6, A → B carries 4, A → T carries 5, B → T carries 8. Send 5 along A → T; A still has 3 spare, so send 3 along A → B; B now receives 6 + 3 = 9 but B → T carries only 8. Total = 5 + 8 = 13. The cut {A → T, B → T} = 13 confirms it; the source cut {S → A, S → B} = 14 is looser and is the standard wrong answer.

  1. For a points table, write the arithmetic of the system first. With 3 for a win and 1 for a draw, a round robin of n teams has n(n − 1)/2 matches, and total points = 3 × (decisive matches) + 2 × (drawn matches) = 3M − D, where D is the number of draws.
  2. Find the number of draws from the points total, then decompose each team's points. 7 points from 3 games is only (2 wins, 1 draw); 3 points is (1 win) or (3 draws) — use the total draws to decide.
  3. Check that draws pair up. Every drawn match gives one draw to each of two teams, so the draws across the table sum to 2D, an even number. An odd sum means a mis-copied cell.

Worked set. Four teams P, Q, R, S play a round robin (6 matches; 3 points for a win, 1 for a draw, 0 for a loss). Final points: P 7, Q 5, R 3, S 1. Total points = 16 = 3 × 6 − D, so D = 2 drawn matches, contributing 4 draw entries across the table. P (7) = 2W 1D 0L. Q (5) = 1W 2D 0L. S (1) = 0W 1D 2L. Draw entries so far 1 + 2 + 1 = 4, so R has no draws and R (3) = 1W 0D 2L. Now the fixtures: Q's two draws must be with the two other teams that drew, P and S. P's two wins are therefore over R and S; Q's win is over R (the only match Q has left); R's win is over S. Every match is now determined: P beat R and S, drew Q; Q beat R, drew P and S; R beat S; and S lost to P and R, drew Q. Points recheck: S = 1 ✓, R = 3 ✓, Q = 3 + 2 = 5 ✓, P = 6 + 1 = 7 ✓.

🧠 Pascal's triangle is the grid
The addition method on an unobstructed grid produces Pascal's triangle turned on its side — 1, 4, 10, 20, 35 along a diagonal — which is why C(m + n, n) works. Knowing that lets you sanity-check a hand count: in a 4 × 3 grid, the junction numbers along the top row must read 1, 4, 10, 20, 35. If yours read 1, 4, 10, 19, 34, you dropped an addition somewhere.

Scheduling and Selection

A scheduling set gives tasks or people, time slots, and rules of two kinds: precedence ("B can start only after A finishes") and exclusion ("P and Q cannot be in the same slot"). Precedence rules form a chain, and the length of the longest chain is the shortest possible completion time — no amount of extra workers shortens it, because each task in the chain must wait for the one before. Exclusion rules are handled exactly like the crosses of an arrangement grid: mark the forbidden pairs and let the eliminations force the slots.

  1. Draw the tasks as nodes with arrows for "must precede". Write each task's duration inside its node.
  2. Compute the earliest finish of each node = its duration + the largest earliest-finish among its predecessors. Work left to right.
  3. The largest earliest-finish is the minimum completion time, and the chain that produced it is the critical path. Any task off that path has slack and can be moved without delaying the end.
  4. For selection with exclusions, count the unrestricted selections and subtract the forbidden ones, or list the valid teams outright when the numbers are small. Listing is safer under time pressure whenever there are fewer than about fifteen cases.

Worked set (precedence). Five tasks: A takes 3 days and has no prerequisite; B takes 2 days and follows A; C takes 4 days and follows A; D takes 1 day and follows both B and C; E takes 2 days and follows D. Earliest finishes: A = 3; B = 3 + 2 = 5; C = 3 + 4 = 7; D = 1 + max(5, 7) = 8; E = 8 + 2 = 10 days. The critical path is A → C → D → E. The wrong answers: 12 (adding every duration, as if one person did everything in sequence), 8 (following A → B → D → E and ignoring that D must also wait for C), and 7 (stopping at the longest pair). B has 2 days of slack: it can start anywhere between day 3 and day 5 without moving the finish.

Worked set (selection with exclusion). A team of 3 is chosen from P, Q, R, S, T. P and Q cannot both be chosen; if R is chosen then S must be chosen. Unrestricted: C(5, 3) = 10. Teams with both P and Q: the third member is R, S or T — 3 teams, leaving 7. Of those 7, teams with R but without S: {P, R, T} and {Q, R, T} — 2 teams (the team {P, Q, R} was already removed). Valid teams = 7 − 2 = 5: PRS, PST, QRS, QST, RST. Listing confirms it, and shows the pattern such sets rely on — the two rules overlap on one team, and a candidate who subtracts 3 and then 3 again gets 4.

ℹ️ Time slots are an arrangement grid with a clock
A set with five speakers and five hourly slots is a linear arrangement: "R speaks immediately after P" is an adjacency, "Q speaks before 11 a.m." is a position bound, and "S and T are not in consecutive slots" is a gap condition. Use the grid method of the previous chapter, with the slots as positions. The only new element is duration — a two-hour talk occupies two consecutive positions, which is a block that moves as one.

Key takeaways

  • A three-set Venn has seven inside regions; write the centre first, then the three two-way regions, then the three "only" regions, and add them to find "none".
  • Exactly two = S₂ − 3T; at least two = S₂ − 2T; exactly one = |A| + |B| + |C| − 2·S₂ + 3T. "A and B" includes the centre unless the set says "but not C".
  • Minimum overlap of two sets is |A| + |B| − N; of three sets it is |A| + |B| + |C| − 2N. Maximum overlap is the smallest set. Construct the extreme case rather than quoting it.
  • A distribution grid is solved by its totals: enter the zeros first, turn relations into row equations, and close each column with its total.
  • Shortest routes m east and n north number C(m + n, n); with a blocked junction, add the numbers west and south at every node instead. Routes through a point multiply; routes avoiding it subtract.
  • In a points table with 3 for a win and 1 for a draw, total points = 3M − D, which gives the number of draws before any match is examined. The minimum completion time of a schedule is its longest dependency chain.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In a colony of 60 households, 35 subscribe to a Hindi newspaper, 30 subscribe to an English newspaper and 12 subscribe to both. How many households subscribe to neither?

    1. 7
    2. 53
    3. 18
    4. 23
    Show answer

    Answer: A — 7

    Households taking at least one paper = 35 + 30 − 12 = 53, because the 12 who take both were counted in each of the 35 and the 30. Neither = 60 − 53 = 7. The value 53 is the union itself — the number who take at least one, not the number who take none. The value 18 is the number taking only the English paper (30 − 12), and 23 is the number taking only the Hindi paper (35 − 12); both are regions of the diagram but neither is the region outside both circles.

  2. Of 100 students, 50 play cricket, 45 play football and 40 play hockey. 20 play cricket and football, 15 play football and hockey, 18 play cricket and hockey, and 8 play all three. How many play exactly two of the three games?

    1. 53
    2. 37
    3. 29
    4. 45
    Show answer

    Answer: C — 29

    Each pairwise figure includes the 8 who play all three, so the three "exactly two" regions are 20 − 8 = 12, 15 − 8 = 7 and 18 − 8 = 10, totalling 29. By formula, exactly two = S₂ − 3T = 53 − 24 = 29. The value 53 is the raw sum of the pairwise intersections, which counts the centre three times over. The value 37 is S₂ − 2T, the number who play at least two games — it includes the 8 who play all three. The value 45 subtracts the centre only once from S₂, which is not any meaningful region. As a check, exactly one = 135 − 106 + 24 = 53, and 53 + 29 + 8 = 90 play at least one game.

  3. In a class of 80 students, 60 passed Mathematics, 50 passed Science and 70 passed English. What is the minimum possible number of students who passed all three subjects?

    1. 0
    2. 30
    3. 50
    4. 20
    Show answer

    Answer: D — 20

    Count the failures: 20 failed Mathematics, 30 failed Science, 10 failed English — 60 failures in all. Each failure removes at most one student from the all-three group, so at least 80 − 60 = 20 students passed everything. The formula gives the same: 60 + 50 + 70 − 2 × 80 = 20. The value 0 assumes the sets can be spread out with no forced overlap, but 60 + 50 + 70 = 180 far exceeds twice the class size, so overlap is unavoidable. The value 30 is the two-set minimum for Mathematics and Science alone (60 + 50 − 80) and ignores English. The value 50 is the maximum, not the minimum — the size of the smallest set.

  4. In a survey, 70% of respondents like tea, 65% like coffee and 10% like neither. What percentage like both tea and coffee?

    1. 35%
    2. 45%
    3. 55%
    4. 25%
    Show answer

    Answer: B — 45%

    Those who like at least one drink are 100 − 10 = 90%. Inclusion–exclusion: both = 70 + 65 − 90 = 45%. The value 35% is 70 + 65 − 100, which forgets the 10% who like neither and treats the union as the whole population — the commonest slip in percentage Venn questions. The value 55% subtracts the "neither" figure from the coffee figure (65 − 10), which has no meaning in the diagram. The value 25% is the only-tea region (70 − 45), a real region but not the one asked for. Check: only tea 25 + only coffee 20 + both 45 + neither 10 = 100.

  5. A city is laid out as a rectangular grid of roads. From her home, Meera's office is 3 blocks east and 2 blocks north. Walking only east or north, how many different shortest routes can she take?

    1. 10
    2. 6
    3. 5
    4. 20
    Show answer

    Answer: A — 10

    Every shortest route is a sequence of 5 steps of which exactly 2 go north, so the count is the number of ways to choose those 2 positions out of 5: C(5, 2) = 10. The addition method agrees — the junction numbers build up 1, 3, 6, 10 along the top row. The value 6 multiplies 3 by 2, as if the eastward and northward choices were independent, but they are steps in one sequence, not two separate choices. The value 5 adds the two distances, which is the length of the route rather than the number of routes. The value 20 is 5!/3!, which counts the 2 northward steps as distinguishable and so double-counts every route.

  6. On a rectangular road grid, point B is 4 blocks east and 3 blocks north of point A, and junction P is 2 blocks east and 1 block north of A. Moving only east or north, how many shortest routes from A to B pass through P?

    1. 35
    2. 18
    3. 17
    4. 9
    Show answer

    Answer: B — 18

    A route through P is a route from A to P followed by a route from P to B, and the two legs are chosen independently, so the counts multiply. A to P: 2 east and 1 north, C(3, 1) = 3 routes. P to B: the remaining 2 east and 2 north, C(4, 2) = 6 routes. Through P: 3 × 6 = 18. The value 35 is C(7, 3), the total number of shortest routes from A to B with no condition about P. The value 17 is 35 − 18, the number of routes that avoid P — the answer to the opposite question, and always printed alongside. The value 9 adds the two legs (3 + 6) instead of multiplying them.

  7. Water flows from source S to tank T through pipes with these hourly capacities: S → A 8 units, S → B 6 units, A → B 4 units, A → T 5 units, B → T 8 units. Flow is only in the direction of the arrows. The maximum flow from S to T per hour is:

    1. 14
    2. 11
    3. 13
    4. 17
    Show answer

    Answer: C — 13

    Everything reaching T must pass through A → T (5) or B → T (8), so the flow cannot exceed 13, and that bound is achievable: send 5 along A → T, pass A's spare 3 through A → B, and B then receives 6 + 3 = 9 of which 8 go on to T. Total 5 + 8 = 13. The value 14 is the capacity out of the source (8 + 6), but the pipes into T cannot carry that much — the tightest cut, not the source, limits the flow. The value 11 ignores the A → B link and sends only 5 through A and 6 through B. The value 17 adds the A → B capacity to the two inflows of T, counting the same water twice.

  8. A project has five tasks. A takes 3 days and can start at once. B takes 2 days and C takes 4 days; both can start only after A finishes. D takes 1 day and can start only after both B and C finish. E takes 2 days and starts after D. With unlimited workers, the minimum number of days to finish the project is:

    1. 12
    2. 10
    3. 8
    4. 7
    Show answer

    Answer: B — 10

    The minimum time is the longest chain of dependencies. Earliest finishes: A = 3; B = 3 + 2 = 5; C = 3 + 4 = 7; D must wait for both B and C, so D = 7 + 1 = 8; E = 8 + 2 = 10. The critical path A → C → D → E takes 10 days and no number of workers can shorten it. The value 12 adds every duration (3 + 2 + 4 + 1 + 2) as if a single worker did the tasks one after another, ignoring that B and C run in parallel. The value 8 follows the chain A → B → D → E and forgets that D also waits for C, which finishes two days after B. The value 7 stops at C's finish and omits D and E altogether.