
Let A and B be two finite sets having m and n elements respectively. Then, find the total number of mappings from A and B.
A \[mn\]
B \[{2^{mn}}\]
C \[{m^n}\]
D \[{n^m}\]
Answer
160.8k+ views
Hint: If there are two non-empty sets X and Y. Then we can say there is mapping from X to Y if there exist a unique image in Y for every element in X.
Complete step by step solution: It is given that, there are two finite sets A and B. There are m elements in set A and n elements in set B. Every element of set A can be assigned to any of elements of B. Hence there are n images for every element of set A.
Let x be any element in A such that \[x \in A\].
So, image of x can be any one of the image of any element of B.
Hence, It can be m images for every element in set A.
So, Total number of mapping from A to B can be calculated as follows,
\[n \times n... \times n(m{\rm{ times) = }}{n^m}\].
Total number of mapping from A to B can be \[{n^m}\].
Option ‘D’ is correct
Note: The common mistake happen by students is they consider as there are n elements in set A and m elements in set B. so mapping is \[nm\]. Which is wrong.
Complete step by step solution: It is given that, there are two finite sets A and B. There are m elements in set A and n elements in set B. Every element of set A can be assigned to any of elements of B. Hence there are n images for every element of set A.
Let x be any element in A such that \[x \in A\].
So, image of x can be any one of the image of any element of B.
Hence, It can be m images for every element in set A.
So, Total number of mapping from A to B can be calculated as follows,
\[n \times n... \times n(m{\rm{ times) = }}{n^m}\].
Total number of mapping from A to B can be \[{n^m}\].
Option ‘D’ is correct
Note: The common mistake happen by students is they consider as there are n elements in set A and m elements in set B. so mapping is \[nm\]. Which is wrong.
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