If two sets A and B have 99 elements in commo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The number of elements common to the Cartesian products $A 	imes B$ and $B 	imes A$ is given by the intersection of the two sets:$n((A 	imes B)

Vedclass · Accepted Answer

$99^2$

Vedclass · Answer

(B) The number of elements common to the Cartesian products $A 	imes B$ and $B 	imes A$ is given by the intersection of the two sets:$n((A 	imes B) \cap (B 	imes A))$Using the property of Cartesian products,$(A 	imes B) \cap (B 	imes A) = (A \cap B) 	imes (B \cap A)$.Since $n(A \cap B) = 99$,we have $n(B \cap A) = 99$.Therefore,the number of elements is $n(A \cap B) 	imes n(B \cap A) = 99 	imes 99 = 99^2$.

If two sets $A$ and $B$ have $99$ elements in common,then the number of elements common to each of the sets $A \times B$ and $B \times A$ is:

Similar Questions

Range of the function $f(x) = 9 - 7\sin x$ is

If $A = \{a, d\}, B = \{b, c, e\}$ and $C = \{b, c, f\},$ then $A \times (B \cup C) =$

In a class,$50$ students play cricket,$20$ students play football and $10$ play both cricket and football. How many students play at least one of these two games?

If $n(U) = 600$,$n(A) = 100$,$n(B) = 200$,and $n(A \cap B) = 50$,then $n(\bar{A} \cap \bar{B})$ is (where $U$ is the universal set and $A$ and $B$ are subsets of $U$).

The largest possible set of real numbers which can be the domain of $f(x) = \sqrt {1 - \frac{1}{x}} $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module