Correct multiple counting systematically

Try

If 12 people belong to one club and 9 to another, with 4 in both, the union has 17 people rather than 21. A shared member contributes twice to the raw sum and must be subtracted once.

Predict

Predict what happens with three overlapping sets. Add individual sizes, subtract all pairwise intersections, then add the triple intersection. A member of all three starts with count 3, falls to 0 after the three subtractions, and needs one final addition.

Experiment

Create membership rows for objects in zero, one, two and three sets. Compute each row’s net contribution under the three-set formula. This small table proves the per-object identity; adding the contributions across any finite universe proves the counting formula.

Understand

Worked count: among integers 1 through 30, multiples of 2 number 15, multiples of 3 number 10 and multiples of both number 5. Their union has 15+10−5=20. The intersection uses divisibility by lcm(2,3)=6, not an unrelated product rule applied blindly.

For three properties A,B,C the formula is |A∪B∪C|=|A|+|B|+|C|−|A∩B|−|A∩C|−|B∩C|+|A∩B∩C|. Pairwise counts include triple members. If a data table instead reports “exactly two,” its entries cannot be inserted as inclusive intersections without conversion.

Invalid arithmetic: subtract the triple intersection as well as the pairs. That leaves triple members with negative net contribution, which cannot count a union.

Intersecting attributes: count sets, not choices of witnesses

Arrange twelve distinct tickets in three rows and four columns. Choose an unordered set of three tickets without replacement. Require at least one ticket from the first row and at least one from the first column. The corner ticket belongs to both marked sets, so “choose a row ticket, choose a column ticket, then choose the rest” can both miss shared-ticket cases and count other sets multiple times.

One ticket can satisfy both requirements

Choose an unordered set of tickets. The first row and first column are marked. Construct an example that meets both, then one that meets both but uses too many marked tickets for the exact-one condition.

Rows 234 Columns 234 Set size 01234 Reset

(1, 1) row column(1, 2) row(1, 3) row(1, 4) row(2, 1) column(2, 2)(2, 3)(2, 4)(3, 1) column(3, 2)(3, 3)(3, 4)

Selected 0/3; marked-row count 0, marked-column count 0. Complete the set before classifying this size-k example.

All size-k subsets	220
Missing marked row	56
Missing marked column	84
Missing both (restore intersection)	20
At least one from each	100
Exactly one: shared intersection ticket	15
Exactly one: two distinct marked tickets	36
Exactly one from each, total	51

Under uniform sampling of unordered size-3 subsets, divide each event count by 220 for its probability. Your manually chosen example is not a random sample. At most16 tickets and1820 subsets; state is temporary and no mastery is awarded.

For an r×s grid and a size-k subset, let a bad event mean missing the marked row and another mean missing the marked column. There are C(rs,k) total subsets, C((r−1)s,k) missing the row and C(r(s−1),k) missing the column. Missing both leaves precisely the (r−1)(s−1) outside tickets. Thus the count meeting both marked sets is C(rs,k) − C((r−1)s,k) − C(r(s−1),k) + C((r−1)(s−1),k). Here C(n,k) counts unordered k-subsets, with value zero when k is negative or greater than n. A set missing both was subtracted twice and must be restored once; a set missing just one has net weight zero; a set meeting both has net weight one. This verifies every sign by membership, not by memorizing a formula.

At r=3,s=4,k=3 the count is220−56−84+20=100. Uniformly choosing an unordered three-ticket subset therefore gives probability100/220=5/11. A probability claim needs this sampling assumption; a person deliberately choosing tickets does not generate that uniform distribution.

Exactly one from each is a different event

Write M=(r−1)(s−1) for the number of tickets outside both marked sets. Split an exact-one set into disjoint cases. If it includes the shared corner ticket, every other ticket must be outside both; there are C(M,k−1) choices. If it excludes that corner, it must contain one of the s−1 marked-row-only tickets and one of the r−1 marked-column-only tickets, with all remaining tickets outside both. This gives (r−1)(s−1)·C(M,k−2) choices. The two cases exhaust the event and never overlap, so add them.

In the twelve-ticket example these counts are C(6,2)=15 and6·C(6,1)=36, totaling51. The same event can be verified by selecting three tickets in the inspector: the shared corner plus two outside tickets, or two distinct marked tickets plus one outside. A set containing the corner and another marked-row ticket satisfies the at-least condition but fails exact-one. At k=1 only the corner works; at k=0 neither event occurs. These endpoints check why blindly requiring two distinct witnesses is wrong.

For a different representation, think of tickets as objects with two attributes, such as rank and suit or product type and warehouse. The counting proof needs distinct objects arranged in a complete grid, unordered selection and no replacement. Changing to ordered draws or allowing repeated objects changes the sample space and requires a new count.

Apply

Check that the claimed set sizes are jointly possible, and distinguish inclusive membership from “only” categories. For a complement such as neither club, subtract the union from the explicitly stated universe size.

Continue

Depth: the general formula alternates signs over nonempty collections of sets. An object belonging to k properties receives total contribution Σ(j=1..k)(−1)ʲ⁺¹C(k,j)=1, following from (1−1)ᵏ=0. This is why the correction pattern continues beyond three sets.