Universal Set
The rectangle \(U\) contains every object under discussion.
Complete notes for SSC Combined Graduate Level examinations
The rectangle \(U\) contains every object under discussion.
The shared lens contains members common to both sets.
Everything in either set, including the overlap.
One class lies wholly inside a broader class.
No member is common; the circles do not overlap.
Members in \(U\) but outside the named set.
Seven internal regions track only, pairwise-only, and all three.
“Some” places at least one definite member in a region.
A set is a well-defined collection. If object \(x\) belongs to set \(A\), write \(x\in A\); otherwise \(x\notin A\).
| Wording | Notation | Region | Counting expression |
|---|---|---|---|
| Both / common | \(A\cap B\) | Overlap only | \(n(A\cap B)\) |
| At least one | \(A\cup B\) | All parts of both circles | \(n(A)+n(B)-n(A\cap B)\) |
| Only \(A\) | \(A-B\) | \(A\) excluding overlap | \(n(A)-n(A\cap B)\) |
| Exactly one | \((A-B)\cup(B-A)\) | Two non-overlap wings | \(n(A)+n(B)-2n(A\cap B)\) |
| Neither | \((A\cup B)'\) | Outside both circles | \(n(U)-n(A\cup B)\) |
The overlap is counted twice in \(n(A)+n(B)\), so subtract it once.
“Only” removes every overlap containing the named set.
The rectangle outside the circles is part of the total.
Every person/object must occupy exactly one final region.
Single totals count the centre three times. Subtracting the three pairwise intersections removes it three times, leaving zero; add it once to count it correctly.
| Required region | Formula |
|---|---|
| \(A\cap B\) only | \(n(A\cap B)-n(A\cap B\cap C)\) |
| Only \(A\) | \(n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)\) |
| Exactly two | \(n(A\cap B)+n(B\cap C)+n(C\cap A)-3n(A\cap B\cap C)\) |
| At least two | \(n(A\cap B)+n(B\cap C)+n(C\cap A)-2n(A\cap B\cap C)\) |
| Exactly one | Sum of the three only-regions |
| None | \(n(U)-n(A\cup B\cup C)\) |
Always work from deepest overlap outward.
Unless “only” is stated, \(n(A\cap B)\) normally includes members also in \(C\). Subtract the triple intersection to obtain pairwise-only.
If neither and all inside regions are known, add them. If total and union are known, subtract to find neither.
| Verbal relationship | Diagram | Example structure |
|---|---|---|
| Every \(A\) is \(B\) | \(A\) inside \(B\) | Squares inside rectangles |
| No \(A\) is \(B\) | Separate circles | Even and odd integers |
| Some \(A\) are \(B\), some are not | Partial overlap | Teachers and writers |
| \(A\) and \(B\) are separate subsets of \(C\) | Two disjoint circles inside a larger circle | Two exclusive species within a genus |
| Three independent categories | Three overlapping circles | Three optional interests |
| Same class under two names | Coincident circles | Equivalent definitions |
Specific class inside broader class.
No possible common member.
Attributes can coexist for some members.
Both descriptions pick exactly the same members.
Place \(S\) entirely inside \(P\). Do not imply that all \(P\) are \(S\).
Use disjoint circles. The relation safely converts: no \(P\) is \(S\).
Place an existence dot in the overlap. The circles need not otherwise be fixed.
A conclusion follows only if it holds in every diagram consistent with the premises.
A possibility follows if at least one valid diagram permits it and no premise forbids it.
| Check | Requirement |
|---|---|
| Non-negativity | Every final region count must satisfy \(x\ge0\) |
| Intersection limit | \(n(A\cap B)\le\min\{n(A),n(B)\}\) |
| Union bounds | \(\max\{n(A),n(B)\}\le n(A\cup B)\le n(A)+n(B)\) |
| Total reconciliation | Sum of mutually exclusive final regions equals \(n(U)\) |
| Integer condition | Counts of discrete members are whole numbers |
| Wording match | “Both” may include triple overlap; “both only” excludes it |
For three sets, fill all three, then pairwise-only, then single-only, then neither.
Two-set union: add both set totals and subtract their common part once.
In three-set inclusion–exclusion: add singles, subtract pairs, add the triple once.
Remove all overlaps touching the named set; restore triple overlap if it was subtracted twice.
Compute the union first, then \(n(\text{neither})=n(U)-n(\text{union})\).
Sum pairwise intersections and subtract the triple three times.
In syllogisms, subject inside predicate—never reverse the container.
A witness dot captures existence without turning “some” into “all.”
Add every mutually exclusive region. It must equal the universe total.
A negative only-region means the data, interpretation or arithmetic is wrong.
Try to draw one valid arrangement. If successful and not forbidden, “may be” follows.
Mark “only,” “at least,” “exactly,” “none,” and “all three” before calculating.
| Need | Formula |
|---|---|
| Two-set union | \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\) |
| Only \(A\) | \(n(A)-n(A\cap B)\) |
| Exactly one of two | \(n(A)+n(B)-2n(A\cap B)\) |
| Neither | \(n(U)-n(A\cup B)\) |
| Three-set union | \(\sum n(\text{single})-\sum n(\text{pair})+n(A\cap B\cap C)\) |
| Exactly two of three | \(\sum n(\text{pair})-3n(A\cap B\cap C)\) |
| At least two of three | \(\sum n(\text{pair})-2n(A\cap B\cap C)\) |
Adding set totals counts common members twice; inclusion–exclusion repairs it.
“In \(A\) and \(B\)” includes the centre unless “but not \(C\)” or “only” is stated.
Only-region counts exclude every overlap, not merely the deepest overlap.
The universe includes members outside every circle.
All \(A\) are \(B\) does not mean all \(B\) are \(A\).
An existence dot represents at least one member, not the entire class.
Universal statements may not establish that a class contains any member.
For exactly two, triple members contaminate all three pair totals and must be removed three times.
Counts cannot be negative; revisit wording or data placement.
One possible diagram proves possibility; every valid diagram is needed for certainty.
Filling single totals first forces repeated corrections; deepest overlap must come first.
Choose relations stated or logically necessary—not stereotypes about the class names.
Deepest, Face pairs, Singles, Around/outside.
The two-set overlap appears twice in the sum of set totals.
Add the three-way centre back once in inclusion–exclusion.
Look inside the rectangle but outside all circles.
The smaller subject class sits inside the predicate class.
The dot remembers existence without exaggerating quantity.
One valid diagram is enough for possibility; “must” needs every door.
All final regions must add up exactly to the universal total.