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Sum
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in [A × B]?
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Solution
It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.
⇒ Number of elements in set B = 3
Number of elements in [A × B]
= [Number of elements in A] × [Number of elements in B]
= 3 × 3 = 9
Thus, the number of elements in [A × B] is 9.
Concept: Cartesian Product of Sets
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Chapter 2: Relations and Functions - Exercise 2.1 [Page 33]
Q 2Q 1Q 3
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NCERT Class 11 Mathematics
Chapter 2 Relations and Functions
Exercise 2.1 | Q 2 | Page 33
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If set A has p no. of elements and set B has q number of elements then the total number of relations defined from set A to set B is 2pq.
Math Formulas
If set A has p no. of elements and set B has q number of elements then the total numberof relations defined from set A to set B is 2pq.
Definition :-For any set A such that n[A] = n
Then number of all relations on A is
As the total number of relations that can be defined a set A to B is the number of possible subsets of A ×B. If n[A] = p and n[B] = q then n[A × B] = pq and the number of subsets of
Example 1 :-If set
Solution :-
Set A has 2 elements and set B has 3 elements then the no. of relation defined from A to B is
Example 2 :-The number of relations defined on set A = {a,b,c,d}.
Solution :-
Set A has 4 elements
No. of relation defined on set A is
Answer
Verified
Hint: A General Function points from each member of "A" to a member of "B". It never has one "A" pointing to more than one "B", so one-to-many is not OK in a function [so something like "f[x] = 7 or 9" is not allowed]But more than one "A" can point to the same "B" [many-to-one is OK].Injective means we won't have two or more "A"s pointing to the same "B". So many-to-one is NOT OK [which is OK for a general function]. As it is also a function one-to-many is not OK But we can have a "B" without a matching "A" Injective is also called "One-to-One".
Complete step by step solution: The Set A has 4 elements and the Set B has 5 elements and we have to find the number of injective mappings.
Let f be such a function.
Now, f takes inputs from set A whereas the output value of f comes from set B.
Using the fact that injective functions are one-one and onto,
\[f[1]\;\]can take \[5\] values,
\[\;f[2]\;\]can then take only \[4\;\]values ,
\[\;f[3]\;\]can
take only \[3\;\]and
\[\;f[4]\;\]only \[2\].
Hence the total number of functions are \[5 \times 4 \times 3 \times 2 = 120\].
Note: In mathematics, an injective function [also known as injection, or one-to-one function] is a function that maps distinct elements of its domain to distinct elements of its co-domain. In simple words, every element of the function’s co-domain is the image of at most one element of its domain.