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Wednesday, 3 August 2011

Math GRE - #2

If S is a non-empty finite set with k elements, then the number of one-to-one functions from S onto S is:
  • k!
  • k^2
  • k^k
  • 2^k
  • 2^{k+1}
The answer is k!. Suppose our set only consisted of {1,2,3}. A function can be defined as a set of tuples that maps each element of the set to another element in the set (since we're looking at functions from S to S). E.g. f could be the function {(1,2), (2,3), (3,1)}. The left element in the tuple is an input to the function, the right element is an output. Note that we will always have three tuples in our function set, with the left element in each tuple being 1, 2, and 3 respectively.

So the question is: how many unique functions can we create by changing the numbers in the tuples? Well, we know the function is one to one, so we know that if our function was {(1, x), (2, y), (3, z)}, x, y, and z have to be different. If we take our example function and choose in the order x, y, z, there are 3 choices for x (out of {1, 2, 3}), 2 choices for y (out of {1, 3}) and only one choice for z (only {1} is left). So the total number of possible functions we could have had is: 3!=3\cdot2\cdot1=6. Thus, for a non-empty finite set with k elements, then the number of one-to-one functions from S onto S is k!.

What if we didn't require the functions to be one-to-one? Which of the answers would it be?

7 comments:

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[hide] sean said...

If we don't require functions to be one-to-one, the answer would be k^k.

I think.

on 4 August 2011 at 15:50
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[hide] Mattie said...

@Sean:

It is k^k if we lose the requirement of injectivity. Each of the k elements of S can be mapped to any element of S, of which there are k.

on 4 August 2011 at 17:54
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[hide] Anonymous said...

Also, since the domain and the codomain are one and the same, would the number of surjections be the same as the number of injections as well?

on 4 August 2011 at 20:56
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[hide] Paul Liu said...

@leegao:

Yes, they would be. Any one to one function would automatically be onto.

on 4 August 2011 at 21:48
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[hide] Mattie said...

@Paul:

Just to qualify a bit more carefully -- any one-to-one function on a finite set is automatically onto

on 5 August 2011 at 08:20
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[hide] Paul Liu said...

@Mattie:

Right. I was thinking in the context of this question, but it never hurts to be too precise.

Thanks!

on 5 August 2011 at 09:11
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[hide] Kevin said...

If you think in terms of group theory, since the injective functions are bijections, the number of bijections is the order of the symmetric group S_k, which is k!

on 6 August 2011 at 06:41
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