what is the number of injective functions from $f: \mathbb{Z}/5 -> \mathbb{Z}/8$ with $f(1)=0$ ?
The forumula is $n!/(n-m)! $
My solution:
With f(1) = 0 we have 4 for m and 7 for n. As result I get $7!/(7-4)!=840$
Is that correct?
Number of injective functions
1
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combinatorics
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0The question is, unfortunately, not understandable in its current form. What is $n$, $m$, $\Bbb Z/5$, $\Bbb Z/8$? If you want to map a set of $4$ elements into a set of $7$ elements injectively, there are $7\times 6\times 5\times 4$ possibilities, if you ask for this. – 2017-02-06
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0By $\mathbb Z/5$, do you mean $\mathbb Z_5$? – 2017-02-06
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0@PeterFranek : which is exactly the answer he gave. Please try to consider not all users are americans, and there are different ways to note quotient groups : in France, we use $\mathbb Z/n\mathbb Z$, or $\mathbb Z/(n)$... You understood the question, no need to give a bad mark to the author... He showed you efforts of research, after all :-) – 2017-02-06
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0@NicolasFRANCOIS Ok; I didn't downvote, by the way – 2017-02-06
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0@PeterFranek : I didn't say you did ;-) – 2017-02-06