How many one-to-one functions are there from a set with 5 elements to a set with 4 elements
$\begingroup$ Show Consider functions from a set with $5$ elements to a set with $3$ elements. a) Each element mapped to $3$ images. b) $0$ c) How do I do this? Edit: I tried doing this way. EDIT: There can be a set of cardinality {3,1,1} or {2,2,1}. For {3,1,1}: 5C3 * 2C1 * 1C1 * 3! For {2,2,1}: 5C2 * 3C2 * 1C1 * 3! And i realized my 3! is wrong. Should be * 3 only. Why is that so? asked Apr 28, 2016 at 8:47
RStyleRStyle 6071 gold badge5 silver badges14 bronze badges $\endgroup$ 5 $\begingroup$ You correctly found that there are $3^5$ functions from a set with five elements to a set with three elements. However, this counts functions with fewer than three elements in the range. We must exclude those functions. To do so, we can use the Inclusion-Exclusion Principle. There are $\binom{3}{1}$ ways of excluding one element in the codomain from the range and $2^5$ functions from a set with five elements to the remaining two elements in the codomain. There are $\binom{3}{2}$ ways of excluding two elements in the codomain from the range and $1^5$ functions from a set with five elements to the remaining element in the codomain. By the Inclusion-Exclusion Principle, the number of surjective (onto) functions from a set with five elements to a set with three elements is
answered Apr 28, 2016 at 9:11
N. F. TaussigN. F. Taussig 68.2k13 gold badges52 silver badges70 bronze badges $\endgroup$ 2 $\begingroup$ Hint on c) The "onto"-function will induce a partition of its domain (as any function) and this partition (actually the fibres of the function) will - because it is onto - have exactly $3$ elements. So to be found is in the first place how many such partitions exist. A fixed partition gives room for $3\times2\times1=6$ functions. So you end up with: $$6\times\text{number of partitions on }\{1,2,3,4,5\}\text{ that have exactly }3\text{ elements}$$ Also have a look here (especially the counting of partitions). A general formula for the number of onto-functions $\{1,\dots,n\}\to\{1,\dots,k\}$ is: $$k!S(n,k)$$where $S(n,k)$ stands for the Stirling number of the second kind. answered Apr 28, 2016 at 8:55
drhabdrhab 143k9 gold badges70 silver badges191 bronze badges $\endgroup$ 1 How many one(d) 2520 one-to-one functions.
How many oneTherefore, there are one-to-one functions from the set with 5 elements to the set with 4 elements.
How many onto functions are there from a set with 5 elements to a set with 3 elements?1 Answer. Image of each element of A can be taken in 3 ways. ∴ Number of functions from A to B = 35 = 243.
How many oneThe number of one to one functions is N!, because the max mapping to Y is N.
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