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What a set is, and when two sets are the same
Here are two descriptions. The first is the set $\{2, 4, 6\}$. The second is “the set of even numbers between 1 and 7”. They look nothing alike. Are they two sets, or one?
Work the second one out. The even numbers above 1 and below 7 are 2, 4 and 6. Nothing else qualifies. So the second description holds 2, 4 and 6 — exactly what the first one lists. One set, written down two ways.
That is the rule this chapter rests on. A set is a well-defined collection of distinct objects, and two sets are equal exactly when they hold the same elements. The wording used to name a set does not enter into it. Membership decides identity, and nothing else does.
The words “well-defined” are doing real work too. “The set of tall students” is not a set, because nobody can say for certain who belongs. “The set of students above 170 cm” is a set, because every student either qualifies or does not.
Ask what belongs, never how it was described.
Writing a set down: roster and set-builder
A set can be written down in two ways, and both are standard.
ROSTER form lists every element inside braces, separated by commas: $\{1, 3, 5, 7, 9\}$. You can see at a glance exactly what is in it. That works well while the set stays small.
SET-BUILDER form states the property instead of listing members: $\{x : x = 2 n - 1, n \in \mathbb{N}, n \leq 5\}$. Read it aloud as “the set of all x such that x is one less than twice n”. The conditions after the comma say what n may be: a natural number, and no bigger than 5. Work through n = 1, 2, 3, 4, 5 and you get 1, 3, 5, 7, 9.
Both forms name the same set — the first five odd natural numbers. Neither is more correct than the other. Roster is clearer for a small set. Builder is the only practical choice once a set is large, or infinite.
Two notations, one set. Choose by size, not by rule.
Convert $\{x : x^2 < 20, x \in \mathbb{N}\}$ to roster form
- $1^2 = 1$, $2^2 = 4$
test the natural numbers in order, starting from the smallest — both are under 20, so both qualify - $3^2 = 9$, $4^2 = 16$
continue testing — both still under 20, so both qualify - $5^2 = 25$
not under 20 — stop; squares grow as the numbers grow, so every natural number past 5 has a square larger than 25 too - $\{1, 2, 3, 4\}$
the roster form — every candidate tested before the condition first failed
Empty, finite, infinite
A set with no elements at all is the EMPTY SET, written $\emptyset$ or $\{\}$. It is an ordinary set in every other respect. It simply has nothing in it.
Watch one thing carefully here. $\{\emptyset\}$ is NOT the empty set. It is a set holding one element, and that element happens to be the empty set. A box with an empty box inside it is not an empty box.
A set is FINITE when counting its elements comes to an end. $\{1, 3, 5, 7, 9\}$ is finite, with 5 elements. The empty set is finite too, with 0 elements.
A set is INFINITE when the counting never ends. $\mathbb{N}$, the natural numbers, is infinite. So is $\mathbb{R}$, the real numbers.
Empty is a size, not a special case. A set holding the empty set has size 1.
Subsets, and the intervals they name on the real line
Set A is a SUBSET of set B when every element of A is also an element of B. We write $A \subseteq B$.
Nothing at all is asked of B beyond that. B may hold extra elements A does not have, or it may hold exactly the same ones.
When B does hold something extra, so that $A \neq B$, we call A a PROPER subset and write $A \subset B$.
Two facts drop straight out of the definition, and both get used constantly. Every set is a subset of itself, $A \subseteq A$, because every element of A is of course in A. And the empty set is a subset of every set, $\emptyset \subseteq A$. It holds no element that could fail the test.
Subset asks one question of every element of A: is it in B as well?
Show $A = \{2, 4\}$ is a proper subset of $B = \{1, 2, 3, 4, 5\}$
- $2 \in B$, $4 \in B$
check each element of A in turn — both pass, so $A \subseteq B$ - $A \neq B$
B holds 1, 3, 5, which A lacks — the two sets are not equal - $A \subset B$
A is a subset of B, and B has an extra element A lacks — together these make A a PROPER subset. Establish the subset first, then look for something extra: both together give proper
Between any two real numbers sit infinitely many more. So a stretch of the real line can never be written in roster form. It gets its own notation instead.
An INTERVAL is the set of every real number between two endpoints. What changes from one interval to the next is whether the endpoints themselves are counted in.
$(a, b)$ is OPEN and leaves both endpoints out: $\{x \in \mathbb{R} : a < x < b\}$. $[a, b]$ is CLOSED and takes both in: $\{x \in \mathbb{R} : a \leq x \leq b\}$. The round bracket shuts an endpoint out. The square bracket lets it in.
Half-open intervals take one of each. $[a, b)$ includes a but not b. $(a, b]$ includes b but not a.
The shape of the bracket, not the letter beside it, tells you whether an endpoint belongs.
The power set — every subset at once
Given a set, you can ask a larger question: what are ALL of its subsets? Collect them into one set and you have the POWER SET of A, written $P(A)$.
Two of those subsets are easy to forget, and both belong. $\emptyset$ is a subset of A, so it is an element of $P(A)$. And A is a subset of itself, so A is an element of $P(A)$ too.
The size follows a clean rule. If A has n elements, then $P(A)$ has exactly $2^n$ elements. Here is where that comes from. To build one subset, walk through A’s elements and make a single decision about each: in, or out. That is 2 choices, made n times, and each choice is independent of the others. So there are $2^n$ combinations, and every combination gives a different subset.
Every subset is one in-or-out decision per element. Count the decisions and you have counted the subsets.
List the power set of $A = \{1, 2, 3\}$
- $\emptyset$
one subset with no elements - $\{1\}$, $\{2\}$, $\{3\}$
three subsets with one element - $\{1,2\}$, $\{1,3\}$, $\{2,3\}$
three subsets with two elements - $\{1,2,3\}$
one subset with all three elements — work up by size, so nothing is missed - $P(A)$: 8 subsets in all
collecting the four groups above — one empty, three with one element, three with two, one with all three - $1 + 3 + 3 + 1 = 8 = 2^3$
check the count against the rule — the list and the formula agree; listing by size catches every subset once and hands you a count to check against
Fixing the universe, and drawing it
Some questions only make sense once you have fixed what is being talked about. “How many students did not pass?” cannot be answered until you know which students are in the discussion.
The UNIVERSAL SET, written $U$, is that fixed collection: everything under discussion in a given problem. Every other set in the problem is understood as a subset of U.
U is not fixed once and for all. In one problem U might be $\mathbb{N}$. In another it might be the students in one class, or the days of one week. You choose it to fit the problem in front of you.
Because it is a choice, it is worth stating plainly at the start of a problem. Change U and the answers change with it.
U is a decision you make, not a fact about sets.
Relationships between sets are easier to see than to read. A VENN DIAGRAM is how they get drawn.
The universal set U is a rectangle. Each set inside it is a closed curve, usually a circle. An element sits inside a curve when it belongs to that set, and outside the curve when it does not.
Where two curves overlap, that region holds the elements both sets share. Where a curve sits on its own, that region holds what only that set has. The space inside the rectangle but outside every curve holds whatever belongs to none of them.
The correspondence is exact, and that is what makes the picture worth trusting. Every region of the diagram stands for one membership combination, and every membership combination has exactly one region.
Region and membership line up one to one. Read the diagram as a claim, not a sketch.
The four operations
The UNION of A and B is written $A \cup B$. It holds every element that is in A, in B, or in both.
Nothing is left out. Every element of A goes in, and so does every element of B.
Nothing is doubled either. If an element sits in both sets, it still appears once in the union — a set never lists the same element twice. That is not a special rule invented for unions. It is what “distinct objects” already meant in the definition of a set.
On a Venn diagram, the union is both circles shaded, overlap included.
Union collects everything. A shared element is collected once, not twice.
The INTERSECTION of A and B, written $A \cap B$, holds the elements that are in BOTH A and B.
Union asked “in either one?”. Intersection asks “in both?”, so it is always the stricter test. An intersection can never come out larger than either set it was built from.
Sometimes nothing passes at all. If A and B share no elements, then $A \cap B = \emptyset$, and A and B are called DISJOINT.
On a Venn diagram, disjoint sets are two circles drawn so they do not touch. There is no overlap region, because there would be nothing to put in it.
Intersection keeps only what both sets agree on — and sometimes they agree on nothing.
Find $A \cup B$ and $A \cap B$ for $A = \{1,2,3,4\}$, $B = \{3,4,5,6\}$
- $A \cup B = \{1, 2, 3, 4, 5, 6\}$
take everything from either set — 3 and 4 turn up in both, so each is listed once, not twice - $A \cap B = \{3, 4\}$
keep only what is in both sets — check each element of A against B in turn - $n(A) = 4$, $n(B) = 4$
compare the sizes — each set on its own has 4 elements - $n(A \cup B) = 6$
not 8 — the two shared elements were counted twice in $4 + 4$, and the union lists them once each; the size of a union is not the two sizes added, unless the sets share nothing
The DIFFERENCE of A and B, written $A - B$, holds the elements of A that are NOT in B.
Read it as “A, with anything B also has taken out”. B’s own elements never enter the result. Only A’s do.
Order matters here, and this is where the mistakes happen. $A - B$ and $B - A$ are generally different sets. $A - B$ keeps A’s leftovers, and $B - A$ keeps B’s. There is no reason those two collections would match.
Compare that with union and intersection, where swapping the two sets changes nothing at all. Difference does not behave that way.
*$A - B$ starts from A. Swap the letters and you have started somewhere else.*
The COMPLEMENT of A, written $A'$, holds everything in the universal set that is not in A. In the notation already built, $A' = U - A$.
So a complement is just a difference, with U always on the left.
That makes the universal set load-bearing rather than decorative. “Not in A” means nothing until “everything under discussion” has been settled. Ask for the complement of the set of prime numbers and the honest reply is a question: complement inside what?
Fix U as the numbers 1 to 10 and one answer follows. Fix U as $\mathbb{N}$ and a completely different one does.
A complement is only as well-defined as the U it is taken inside.
Find $A - B$, $B - A$, and $A'$ for $A = \{1,\ldots,5\}$, $B = \{4,\ldots,7\}$, $U = \{1,\ldots,10\}$
- $A - B = \{1, 2, 3\}$
start with A and drop anything B also has — B has 4 and 5, so both go - $B - A = \{6, 7\}$
start with B this time and drop anything A also has — A has 4 and 5, so those two go again - $A - B \neq B - A$
the two results are different sets, confirming difference is not symmetric on this pair - $A' = \{6, 7, 8, 9, 10\}$
take everything in U that is not in A — U runs 1 to 10, A holds 1 to 5; difference starts from whichever set is written first, complement always starts from U
Laws the operations obey
Set operations obey fixed laws, much as arithmetic does. Knowing them lets you rewrite an expression instead of computing it from the ground up.
Union and intersection are each COMMUTATIVE, so order does not matter: $A \cup B = B \cup A$. Both are ASSOCIATIVE too, so when the same operation is repeated, the grouping does not matter either.
Each one DISTRIBUTES over the other. Union distributes over intersection, and intersection distributes over union. That is one more than arithmetic gives you, where multiplication distributes over addition but addition does not distribute over multiplication.
DE MORGAN’S LAWS connect the complement to both operations, and they flip the operation on the way through. The first says $(A \cup B)' = A' \cap B'$. The second says $(A \cap B)' = A' \cup B'$. The complement of a union is the intersection of the complements, and the other way round.
De Morgan’s laws swap the operation, not just the sets. Union goes in, intersection comes out.
Verify $(A \cup B)' = A' \cap B'$ on one concrete pair
- $A \cup B = \{1, 2, 3, 4, 5, 6, 7\}$
left side, step 1 — compute the union first - $(A \cup B)' = \{8, 9, 10\}$
left side, step 2 — take the complement of the union inside U - $A' = \{6, 7, 8, 9, 10\}$
right side, step 1 — compute the complement of A - $B' = \{1, 2, 3, 8, 9, 10\}$
right side, step 1 continued — compute the complement of B, without looking at the left side - $A' \cap B' = \{8, 9, 10\}$
right side, step 2 — intersect the two complements, keeping only what appears in both - $\{8, 9, 10\} = \{8, 9, 10\}$
both sides match — the law holds on this pair; one concrete pair cannot prove a law, but it can catch a misremembered one
Counting without double-counting
Write $n(A)$ for the number of elements in A. For finite sets, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
The subtraction is the part that matters, so it is worth seeing where it comes from.
Add $n(A)$ and $n(B)$, and each element gets counted once for every set it belongs to. An element in only A is counted once, which is right. An element in both is counted twice. It is counted once inside $n(A)$, then again inside $n(B)$. That is one time too many. Subtracting $n(A \cap B)$ removes exactly one copy of each of those.
When A and B are disjoint there is nothing shared, so $n(A \cap B) = 0$ and the formula collapses to plain addition.
Add, then correct. The correction is the size of the overlap, subtracted once.
Count how many of 50 students play cricket or football
- $n(C) = 30$, $n(F) = 20$
name the sets first — C for cricket players, F for football players — and write down what you were given - $n(C \cap F) = 10$
10 students play both games - $n(C \cup F) = 30 + 20 - 10 = 40$
cricket or football is the union, so the inclusion-exclusion formula applies - $50 - 40 = 10$
the class has 50 students in all, so those playing neither are everyone outside the union
One question, five moves
Every idea in this chapter answers one question: exactly which objects belong to this set?
The five moves are all ways of answering it.
- REPRESENTATIONS — roster and set-builder — are two ways of stating an answer.
- SUBSETS and the POWER SET compare answers across sets and count them.
- The four OPERATIONS build new sets by combining answers.
- The COUNTING FORMULA turns an answer into a tally that never counts a shared member twice.
The real skill this chapter builds is not any one of those five on its own. It is reading a problem and knowing which of them it is asking for.
A question about what fits inside what is a subset question. A question about combining sets is an operation. A question with “how many” in it, and any overlap at all, is the counting formula.
Five moves, one question underneath all of them.
Two traps this chapter sets
THE TRAP. $\emptyset$ and $\{\emptyset\}$ are the same thing, because both are just “nothing”.
THE REALITY. They are never the same set.
$\emptyset$ has 0 elements. $\{\emptyset\}$ has 1 element, and that element is the empty set itself. Count them and the difference is immediate: 0 against 1.
Two sets are equal only when they hold the same elements. A set with nothing in it can never pass that test against a set with one thing in it. What the one thing happens to be makes no difference.
The braces are doing real work in $\{\emptyset\}$. Braces wrap something up and make it a member. Wrapping nothing in braces gives you a container holding one item, not an empty container.
Cardinality 0 and cardinality 1 are never equal. That settles it, with no further argument needed.
Weaker. Asked for $n(\{\emptyset\})$, a student writes 0, and the reasoning sounds reasonable said out loud: the braces hold the empty set, the empty set is nothing, so there is nothing in there to count. The same student then writes $\{\emptyset\} = \emptyset$, on the grounds that two sets that are both “nothing” have to be the same set.
Stronger. Count what is actually listed between the outer braces. One thing is listed: $\emptyset$. That the listed thing is itself empty changes nothing about the count, so $n(\{\emptyset\}) = 1$. And $n(\emptyset) = 0$, so the two sets have different sizes and cannot be equal. The test to carry away is the comparison with $\{0\}$: nobody writes $n(\{0\}) = 0$ because the element happens to be zero, and $\emptyset$ sitting inside braces is no different.
THE TRAP. $n(A \cup B) = n(A) + n(B)$, always. Add the two sizes and you are done.
THE REALITY. That works only when A and B are DISJOINT.
The moment the two sets overlap, plain addition fails. Every element of $A \cap B$ has already been counted once inside $n(A)$. Then it gets counted again inside $n(B)$. Nothing in the addition removes that duplicate.
The fix is one subtraction: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$. Take off one copy of each shared element and the count comes right.
So the safe habit is to look for an overlap before adding. If the overlap really is empty, then $n(A \cap B) = 0$ and the subtraction costs you nothing.
Adding the sizes is not wrong, it is incomplete. It is the disjoint case of a formula that has one more term.
Practice set
- practice Write $\{x : x \in \mathbb{N}, x^2 < 30\}$ in roster form.
- practice Write $\{2, 4, 6, 8, 10\}$ in set-builder form.
Answers
- $\{1, 2, 3, 4, 5\}$
- $\{x : x = 2n, n \in \mathbb{N}, n \leq 5\}$
- practice Is $\{x \in \mathbb{N} : x < 1\}$ empty, finite, or infinite?
- practice Classify $\{x \in ZZ : x > 5\}$ as finite or infinite.
Answers
- Empty (hence finite, 0 elements).
- Infinite.
- practice Which of these is NOT a subset of $\{1, 2, 3\}$?
- practice Write $\{x \in \mathbb{R} : -2 \leq x < 5\}$ using interval notation.
- practice If $A = \{a, b, c\}$, how many elements does $P(A)$ have?
- practice How many elements does $P(A)$ have if $n(A) = 6$?
Answers
- $\{1,2,3,4\}$
- $[-2, 5)$
- $8$
- $2^6 = 64$
- practice Universal set $U = \{1,2,\ldots,10\}$. If $A = \{1,3,5,7,9\}$ is drawn as a circle inside U’s rectangle, which numbers lie inside the rectangle but outside the circle?
- practice In a Venn diagram of two sets A and B inside universal set U, which region represents $(A \cup B)'$?
- practice In a Venn diagram of two sets A and B inside universal set U, which region represents $A - B$?
Answers
- $\{2,4,6,8,10\}$
- inside the rectangle but outside both circles
- inside A only, not B
- practice Let $A = \{1,2,3,4\}$, $B = \{3,4,5,6\}$. Find $A \cup B$ and $A \cap B$.
- practice Universal set $U = \{1,\ldots,10\}$, $A = \{1,2,3,4,5\}$, $B = \{4,5,6,7\}$. Find $A - B$ and $B - A$.
- practice Using De Morgan’s laws, $(A \cup B)'$ equals which of these?
Answers
- $A \cup B = \{1,2,3,4,5,6\}$; $A \cap B = \{3,4\}$
- $A-B = \{1,2,3\}$; $B-A = \{6,7\}$
- $A' \cap B'$
- practice In a survey of 60 people, 25 like tea, 20 like coffee, and 10 like both. How many like tea or coffee?
- practice If $n(A) = 15$, $n(B) = 10$, $n(A \cap B) = 5$, find $n(A \cup B)$.
- practice If $n(A) = 15$, $n(B) = 12$, $n(A \cap B) = 5$, and $n(U) = 30$, find $n((A \cup B)')$.
Answers
- $35$
- $20$
- $8$
- In a Venn diagram, what does the region where two circles overlap represent?
Check your answer
- A marketing report draws two overlapping circles — one for customers who bought Product A, one for Product B. A customer who bought only Product A appears where on the diagram?
Check your answer
- A Venn diagram compares students who answered Question 1 correctly against students who answered Question 2 correctly, drawn as two overlapping circles inside a rectangle. What does the rectangle itself represent?
Check your answer
Chapter-end problems
- board-easy Write $\{x : x \in \mathbb{N}, x < 6\}$ in roster form.
- board-easy Are $\{2, 4, 6\}$ and $\{6, 4, 2\}$ equal as sets?
- board-easy Is $\{\}$ the same set as $\{0\}$?
- board-easy Let $A = \{1, 2, 3\}$, $B = \{2, 3, 4, 5\}$. Find $A \cup B$.
- board-easy Let $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5, 6\}$. Find $A \cap B$.
- board-easy Let $A = \{1, 2, 3, 4, 5\}$, $B = \{3, 4\}$. Find $A - B$.
- board-easy Universal set $U = \{1, \ldots, 10\}$, $A = \{2, 4, 6, 8, 10\}$. Find $A'$.
- board-easy Is $\{1, 2\}$ a subset of $\{1, 2, 3\}$?
- board-easy If $A = \{a, b, c\}$, how many elements does $P(A)$ have?
- board-easy Which of these sets is infinite?
- board-easy Write $\{1, 4, 9, 16, 25\}$ in set-builder form.
- board-standard $U = \{1, \ldots, 20\}$. $A$ is the multiples of 3 in U, $B$ the multiples of 4. Find $A \cap B$.
- board-standard Let $A = \{x \in ZZ : -3 \leq x \leq 3\}$. Find $n(A)$.
- board-standard $U = \{1, \ldots, 10\}$, $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5, 6\}$. Verify $(A \cup B)' = A' \cap B'$.
- board-standard Let $A = \{1, 2, 3\}$, $B = \{2, 3, 4\}$. Find $(A - B) \cup (B - A)$.
- board-standard In a class of 50, 30 play cricket, 20 play football, and 10 play both. How many play neither?
- board-standard In a Venn diagram of sets A and B inside U, which region is $A \cap B'$?
- board-standard List every element of $P(\{1, 2\})$.
- board-standard Write the interval $[2, 7)$ as a set using set-builder notation.
- board-standard If $A$ is a subset of $B$, and $B$ is a subset of $C$, does $A$ have to be a subset of $C$?
- board-standard A school has 50 students. 18 play chess; the rest do not. Find $n(A')$ where $A$ is the chess players.
- board-standard $U = \{1, \ldots, 10\}$, $A = \{1, 3, 5, 7, 9\}$. Verify $(A')' = A$.
- board-standard In a survey of 100 people, 55 read newspaper A, 40 read B, and 15 read both. How many read A or B?
- JEE In a survey: 50 like tea, 40 like coffee, 30 like juice; 15 like tea & coffee, 10 coffee & juice, 8 tea & juice, and 5 like all three. How many like at least one?
- JEE If $n(P(A)) = 64$, find $n(A)$.
- JEE Let $A = \{1,2,3,4,5\}$, $B = \{2,3\}$, $C = \{4\}$. Verify $A - (B \cup C) = (A - B) \cap (A - C)$.
- JEE Let $A = \{x : x^2 - 5x + 6 = 0\}$ and $B = \{2, 3, 5\}$. Is $A$ a subset of $B$?
- JEE In a group of 70 people, everyone likes tea or coffee (or both); 45 like tea, 35 like coffee. How many like both?
- JEE If $A = \{1, 2, 3\}$ and $B = \{4, 5\}$ are disjoint, which formula gives $n(A \cup B)$?
- JEE $U = \{1,\ldots,10\}$, $A=\{1,2,3,4,5\}$, $B=\{2,3,4,5,6\}$, $C=\{3,4,5,6,7\}$. Verify $(A \cap B \cap C)' = A' \cup B' \cup C'$.
- JEE A set has 5 elements. How many of its subsets contain one fixed element $x$?
- JEE In a survey of three magazines A, B, C: 50 read A, 40 read B, 30 read C; 15 read A&B, 10 read B&C, 8 read A&C, 5 read all three. How many read ONLY C?
Answers
- $\{1, 2, 3, 4, 5\}$
- Yes — order never changes a set’s identity.
- No — they have different cardinalities.
- $\{1, 2, 3, 4, 5\}$
- $\{3, 4\}$
- $\{1, 2, 5\}$
- $\{1, 3, 5, 7, 9\}$
- Yes.
- $8$
- $\mathbb{N}$, the natural numbers
- $\{x : x = n^2, n \in \mathbb{N}, n \leq 5\}$
- $\{12\}$
- $7$
- $\{7, 8, 9, 10\}$, and both sides agree.
- $\{1, 4\}$
- $10$
- inside A only, not B
- $\{\{\}, \{1\}, \{2\}, \{1,2\}\}$
- $\{x \in \mathbb{R} : 2 \leq x < 7\}$
- Yes — the subset relation is transitive.
- $32$
- They are equal — double complement returns A.
- $80$
- $92$
- $6$
- $\{1, 5\}$ on both sides — the identity holds.
- Yes.
- $10$
- $n(A) + n(B)$
- $\{1, 2, 6, 7, 8, 9, 10\}$ on both sides.
- $16$
- $17$
JEE-level practice (item bank)
- If n(A) = 20, n(B) = 30 and n(A ∪ B) = 40, find n(A ∩ B).
Check your answer
- In a survey of 100 students, 45 play cricket, 30 play football, and 20 play both. How many play neither?
Check your answer
- If n(P(A)) = 32, find n(A).
Check your answer
- How many PROPER subsets does a set with 4 elements have?
Check your answer
- A = {1,2,3,4,5,6}, B = {2,4,6,8}. Find n(A-B).
Check your answer
- 18 students play chess, 20 play carrom, 27 play table tennis; 7 play chess and carrom, 12 play carrom and table tennis, 4 play chess and table tennis, and 2 play all three. How many play at least one game?
Check your answer
- If A ⊆ B, which of these is always true?
Check your answer
- How many elements does P(∅), the power set of the empty set, have?
Check your answer
- n(A) = 15, n(B) = 25, disjoint, n(U) = 50. Find n((A ∪ B)′).
Check your answer
- n(A) = 12, n(B) = 18, n(A ∩ B) = 5. Find n(A Δ B), the symmetric difference of A and B.
Check your answer
- If n(A) = n(B) and n(A ∩ B) = n(A), what follows?
Check your answer
- How many subsets of {1, 2, ..., 10} contain neither 1 nor 2?
Check your answer
- In a town of 10,000 families, 40% read newspaper A, 20% read B, and 10% read both. What percentage reads neither?
Check your answer
- A = {x : x is a prime number, x < 20}. Find n(A).
Check your answer
- Which of these correctly expresses (A - B)′ (relative to universal set U)?
Check your answer
- 70 people like tea, coffee, or both; 45 like tea, 35 like coffee. How many like both?
Check your answer
- If n(A ∪ B ∪ C) = n(A) + n(B) + n(C), what must be true?
Check your answer
- A set has 5 elements. How many of its subsets contain one fixed element x?
Check your answer
- How many 2-element subsets does a 6-element set have?
Check your answer
- In a survey of three magazines A, B, C: 50 read A, 40 read B, 30 read C; 15 read A and B, 10 read B and C, 8 read A and C, and 5 read all three. How many read ONLY C?
Check your answer