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Let A and B be finite sets. Let a be the size of A. Let b be the size of B. Assume 0 < a < b.

(a) How many functions are there with domain A and co-domain B?
(b) How many one-to-one functions are there with domain A and co-domain B?
(c) How many one-to-one functions are there with domain B and co-domain A?
(NOTE- domain is B, co-domain is A.)
(d) How many onto functions are there with domain A and co-domain B?

How are we supposed to figure out how many functions there are? Couldn't it be pretty much infinite since each function can do things differently and still get the same value?

How would we solve a)?

Would both b) be a or !b/!a? and would c) be 0?

I understand that d) should be 0, but what if the domain was B and co-domain A? how would we solve that then?

-thanks and any explanation would be appreciated

Mark
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    Functions are uniquely determined by their values, so if functions have the same values on all elements of the domain, they cannot "do things differently". – Tobias Kildetoft Dec 17 '14 at 09:05
  • @TobiasKildetoft Ah okay I didn't know thats how they were defined. so would a) be B? – Mark Dec 17 '14 at 09:08
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    @Mark There are $b^a$ number of functions. – Rustyn Dec 17 '14 at 09:09
  • or would it be more complicated than that. Like B^B – Mark Dec 17 '14 at 09:09
  • @Rustyn oh okay. I understand that now. B different mappings for each element of A. so b * b * b. thanks – Mark Dec 17 '14 at 09:10
  • @Mark sure no problem. – Rustyn Dec 17 '14 at 09:11
  • @Rustyn could you double check me on b) and c) and possibly elaborate on how we would solve d) if the codomains and domains were switched? Sorry if that's asking for a lot. I'm trying to get a good grasp on these concepts – Mark Dec 17 '14 at 09:13
  • @Mark, You may find the following and the link therein helpful. http://math.stackexchange.com/questions/334420/number-of-onto-functions – Radz Dec 17 '14 at 10:33
  • This may also be helpful. http://math.stackexchange.com/questions/223240/how-many-distinct-functions-can-be-defined-from-set-a-to-b – Radz Dec 17 '14 at 10:35

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