Set Theory

Suppose $P_1,P_2,\dots,P_{30}$ are thirty sets each having $5$ elements and $Q_1,Q_2,\dots,Q_n$ are $n$ sets with $3$ elements each. Let $\bigcup_{i=1}^{30}P_i=\bigcup_{j=1}^{n}Q_j=S$ and each element of $S$ belongs to exactly $10$ of the $P$’s and exactly $9$ of the $Q$’s. Then $n$ equals

Events $A$ and $B$ satisfy: $P(A \cup B) = \dfrac{1}{6}$, $P(A \cap B) = \dfrac{1}{4}$, $P(A) = \dfrac{1}{4}$ Then events $A$ and $B$ are:

From 50 students: 37 passed Math, 24 Physics, 43 Chemistry. At most 19 passed Math & Physics, at most 29 passed Math & Chemistry, at most 20 passed Physics & Chemistry. Intersection of all 3 is $x$. Find maximum possible value of $x$.

$A_1, A_2, A_3, A_4$ are subsets of $U$ (75 elements). Each $A_i$ has 28 elements. Any two intersect in 12 elements. Any three intersect in 5 elements. All four intersect in 1 element. Find the number of elements belonging to none of the four subsets.

If $P = {(4n - 3n - 1) : n \in N}$ and $Q = {(9n - 9) : n \in N}$, then $P \cup Q$ equals to:

The total number of relations that exist from a set $A$ with $m$ elements into the set $A \times A$ is:

A set contains $(2n+1)$ elements. If the number of subsets that contain at most $n$ elements is $4096$, then the value of $n$ is:

Let $X$ be the universal set for sets $A$ and $B$. If $n(A)=200,;n(B)=300,;n(A\cap B)=100$, then $n(A'\cap B')=300$ provided $n(X)$ is equal to

Find the number of elements in the union of 4 sets A, B, C and D having 150, 180, 210 and 240 elements respectively, given that each pair of sets has 15 elements in common. Each triple of sets has 3 elements in common and $A \cap B \cap C \cap D = \phi$

If the sets A and B are defined as A = {(x, y) | y = 1 / x, 0 ≠ x ∈ R}, B = {(x, y)|y = -x ∈ R} then

A professor has 24 text books on computer science and is concerned about their coverage of the topics (P) compilers, (Q) data structures and (R) Operating systems. The following data gives the number of books that contain material on these topics: $n(P) = 8, n(Q) = 13, n(R) = 13, n(P \cap R) = 3, n(P \cap R) = 3, n(Q \cap R) = 3, n(Q \cap R) = 6, n(P \cap Q \cap R) = 2 $ where $n(x)$ is the cardinality of the set $x$. Then the number of text books that have no material on compilers is

Let $\bar{P}$ and $\bar{Q}$ denote the complements of two sets P and Q. Then the set $(P-Q)\cup (Q-P) \cup (P \cap Q)$ is

The number of elements in the power set P(S) of the set S = {2, (1, 4)} is

In a survey where 100 students reported which subject they like, 32 students in total liked Mathematics, 38 students liked Business and 30 students liked Literature. Moreover, 7 students liked both Mathematics and Literature, 10 students liked both Mathematics and Business. 8 students like both Business and Literature, 5 students liked all three subjects. Then the number of people who liked exactly one subject is

Let $P = \{\theta : \sin\theta - \cos\theta = \sqrt{2}\cos\theta \}$ and $Q = \{\theta : \sin\theta + \cos\theta = \sqrt{2}\sin\theta \}$ be two sets. Then

Suppose A₁, A₂, ... ₃₀ are thirty sets, each with five elements and B₁, B₂, ...., B_n are n sets each with three elements. Let $\bigcup_{i=1}^{30} A_i= \bigcup_{j=1}^{n} Bj= S$. If each element of S belongs to exactly ten of the Ai' s and exactly nine of the Bj' s then n=

If A = { x, y, z }, then the number of subsets in powerset of A is

If $A = \{4^x- 3x - 1 : x ∈ N\}$ and $B = \{9(x - 1) : x ∈ N\}$, where N is the set of natural numbers, then

If A is a subset of B and B is a subset of C, then cardinality of A ∪ B ∪ C is equal to

If X and Y are two sets, then X∩Y ' ∩ (X∪Y) ' is

Suppose $A_1,A_2,\ldots,A_{30}$ are 30 sets each with five elements and $B_1,B_2,B_3,\ldots,B_n$ are n sets (each with three elements) such that  $\bigcup ^{30}_{i=1}{{A}}_i={{\bigcup }}^n_{j=1}{{B}}_i=S\, $ and each element of S belongs to exactly ten of the $A_i$'s and exactly 9 of the $B^{\prime}_j$'s. Then $n=$

If A={1,2,3,4} and B={3,4,5}, then the number of elements in (A∪B)×(A∩B)×(AΔB)

In a class of 100 students, 55 passed in Mathematics and 67 passed in Physics.The number of students who passed in Physics only is:

Let A and B be sets. $A\cap X=B\cap X=\phi$ and $A\cup X=B\cup X$ for some set X, relation between A & B

Let R be reflexive relation on the finite set a having 10 elements and if m is the number of ordered pair in R, then

The negation of $\sim S\vee(\sim R\wedge S)$ is equivalent to

Given to events A and B such that odd in favour A are 2 : 1 and odd in favour of $A \cup B$ are 3 : 1. Consistent with this information the smallest and largest value for the probability of event B are given by

Let Z be the set of all integers, and consider the sets $X=\{(x,y)\colon{x}^2+2{y}^2=3,\, x,y\in Z\}$ and $Y=\{(x,y)\colon x{\gt}y,\, x,y\in Z\}$. Then the number of elements in $X\cap Y$ is:

Let A and B be two events defined on a sample space $\Omega$. Suppose $A^C$ denotes the complement of A relative to the sample space $\Omega$. Then the probability $P\Bigg{(}(A\cap{B}^C)\cup({A}^C\cap B)\Bigg{)}$ equals

Consider the sample space $\Omega={\{(x,y):x,y\in{\{1,2,3,4\}\}}}$ where each outcome is equally likely. Let A = {x ≥ 2} and B = {y > x} be two events. Then which of the following is NOT true?

Let $A=\{{5}^n-4n-1\colon n\in N\}$ and $B=\{{}16(n-1)\colon n\in N\}$ be sets. Then

Let A = {1,2,3, ... , 20}. Let $R\subseteq A\times A$ such that R = {(x,y): y = 2x - 7}. Then the number of elements in R, is equal to

Suppose that $C$ represents the set of all countries, $R$ represents the set of all countries that have at least one river flowing through it, $M$ represents the set of all countries that have at least one mountain in it, and $D$ represents the set of all countries that have at least one desert in it. It is given that: \[ R \cup M \cup D = C \] Which one of the following gives the set of all countries that have either a mountain or a river, but do not have a desert in it? The notation $D^{c}$ represents the complement of the set $D$ with respect to the universal set $C$.