Which functions are missing from {1,2,3} to {a,b} and why?

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There are 8 possible functions from the set {1,2,3} to {a,b}, calculated using the formula n^m, where n is the number of elements in the codomain and m is the number of elements in the domain. The confusion arose from an initial miscalculation of 6 functions. No injective functions exist in this case, as assigning values to f(1) and f(2) leaves no options for f(3). The discussion confirms that the explanation regarding injective functions is correct. Understanding these concepts is essential for grasping function mappings in set theory.
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Homework Statement


How many functions are there from {1,2,3} to {a,b}? Which are injective? Which are surjective?

Homework Equations


n^m. gives the number of functions

The Attempt at a Solution


To me the number of functions that can be made are 6 because 3x2=6 but I have read online that n^m, n being the elements in the second set and m being the number of elements in the first set, gives you the number of functions. That equation gives me 8 functions but for the life of me I can't figure which functions I'm leaving out.
20160221_200842.jpg

Also there can't be any injective functions because f(1) can take 2 values and f(2) can take one value. Leaving f(3) with none so therefore It can't be injective,correct?
 
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Kingyou123 said:

Homework Statement


How many functions are there from {1,2,3} to {a,b}? Which are injective? Which are surjective?

Homework Equations


n^m. gives the number of functions

The Attempt at a Solution


To me the number of functions that can be made are 6 because 3x2=6 but I have read online that n^m, n being the elements in the second set and m being the number of elements in the first set, gives you the number of functions. That equation gives me 8 functions but for the life of me I can't figure which functions I'm leaving out.
You are missing two functions for which 1 → b .
View attachment 96266
Also there can't be any injective functions because f(1) can take 2 values and f(2) can take one value. Leaving f(3) with none so therefore It can't be injective,correct?
 
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SammyS said:
You are missing two functions for which 1 → b .
Thank you, was my explanation for injective correct?
 
Kingyou123 said:
Thank you, was my explanation for injective correct?
Yes
 
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SammyS said:
Yes
Awesome thank you :)
 

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