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Math Help - Combinatorics question

  1. #1
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    Combinatorics question

    Let A be a set of size n, B a set of size m, C a set of size k. n≤m≤k

    How many ordered pairs of functions (f,g) are there such that f:A-->B, g:B-->C and gf is injective?

    (The working out is what I really want to know).

    Thanks
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  2. #2
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    Hint: if g o f is an injection, then so is f, and g is injective on the image of f. Perhaps you can start by figuring the number of injective f's.
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  3. #3
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    Quote Originally Posted by durrrrrrrr View Post
    Let A be a set of size n, B a set of size m, C a set of size k. n≤m≤k How many ordered pairs of functions (f,g) are there such that f:A-->B, g:B-->C and gf is injective?
    Given |A|=n\le|B|=m\le|C|=k and f:A\to B,~g:B\to C in order for g\circ f:A\to C to be an injection then f must be an injection.
    There are \frac{m!}{(m-n)!} injections A\to B.

    Now for g\circ f:A\to C to be an injection it is not necessary for g to be injective.
    However, g must be injective on the image of f,~f[A].
    How many functions g:B\to C are there such that the restriction to f[A] is an injection?
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  4. #4
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    Quote Originally Posted by Plato View Post
    Given |A|=n\le|B|=m\le|C|=k and f:A\to B,~g:B\to C in order for g\circ f:A\to C to be an injection then f must be an injection.
    There are \frac{m!}{(m-n)!} injections A\to B.

    Now for g\circ f:A\to C to be an injection it is not necessary for g to be injective.
    However, g must be injective on the image of f,~f[A].
    How many functions g:B\to C are there such that the restriction to f[A] is an injection?
    Thank you. There's just one thing I don't understand. This is what I have so far:
    We have m!/(m-n)! injective functions f from A to B.
    We have k!/(k-n)! injective functions g from Im(f) to C.
    Why are we counting the the functions g from B to C that are not in the image of f?
    I know we have k^(m-n) such as these, but why are we counting them?
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  5. #5
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    Quote Originally Posted by durrrrrrrr View Post
    Why are we counting the the functions g from B to C that are not in the image of f? I know we have k^(m-n) such as these, but why are we counting them?
    If we have two mappings g_1~\&~g_2 which agree on f[A] but differ on B\setminus f[A] then is it not true that (g_1, f) ~\&~( g_2, f) are two different pairs? The questions asks for pairs.
    Last edited by Plato; February 24th 2011 at 03:02 AM.
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  6. #6
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    Quote Originally Posted by Plato View Post
    If we have two mappings g_1~\&~g_2 which agree on f[A] but differ on B\setminus f[A] then is it not true that (g_1, f) ~\&~( g_2, f) are two different pairs? The questions asks for pairs.
    Ah. I finally understand. Thank you
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