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 gºf is injective?

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

Thanks

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- Feb 23rd 2011, 08:12 AMdurrrrrrrrCombinatorics 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 gºf is injective?

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

Thanks - Feb 23rd 2011, 09:41 AMemakarov
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.

- Feb 23rd 2011, 09:44 AMPlato
Given $\displaystyle |A|=n\le|B|=m\le|C|=k$ and $\displaystyle f:A\to B,~g:B\to C$ in order for $\displaystyle g\circ f:A\to C$ to be an injection then $\displaystyle f$ must be an injection.

There are $\displaystyle \frac{m!}{(m-n)!}$ injections $\displaystyle A\to B$.

Now for $\displaystyle g\circ f:A\to C$ to be an injection it is not necessary for $\displaystyle g$ to be injective.

**However,**$\displaystyle g$**must be injective on the image of**$\displaystyle f,~f[A].$

How many functions $\displaystyle g:B\to C$ are there such that the restriction to $\displaystyle f[A]$ is an injection? - Feb 24th 2011, 02:20 AMdurrrrrrrr
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? - Feb 24th 2011, 02:43 AMPlato
- Feb 24th 2011, 05:47 AMdurrrrrrrr