Video Transcriptmm. So when you find the number of number of onto functions. Number of onto functions from a set of six elements. From a set of six element to a set of three elements. From a set of six element to a set of three elements. Now uh to solve this we can use to solve this we can use we can use inclusion exclusion principle to solve this problem. We can use inclusion exclusion principle principle. So um for example, so continue the first part for example. So there are 3, 7 that received one ways of excluding that. He's even way of excluding one element, one element in the So basically when the former function uh this uh this other domains and these other in so that you see one ways of excluding one element in the co domain from the range in the core document. From the indeed, what do you mean? The argument from the from the range and from the range and to raise two And tourists to five sorry. and to raise to six and to raise to six function from the set function from the said with six element to the remaining to the demeaning to element And the cover to the remaining two element in the code to me. So similarly uh Three c. Two ways of excluding two element in a court document from a range of 1 to 5 functions from a set of parliament in the remaining elements. So now finally by inclusion including principal, the number of onto function or the subjective function with us. Uh From six element to a set of three elements would be given by three days to six three. to raise to six mile plus three C 2 to raise two, sorry. Plus, It would be one race to 5, 1 days to six. So the final answer would be three days to 6 -3. 7. There were three into to raise to six. Let's try to do is the train three. So that's the final until hope you understand the solution. Thank you. Show
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$\begingroup$ 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 onto functions are there from a set with 6 elements to a set with 4 elements?= 6 and S(4, 3) = 6. Thus, there are 36 onto functions. How many on to functions are there from a set with six elements to a set with three elements?Uh From six element to a set of three elements would be given by three days to six three. to raise to six mile plus three C 2 to raise two, sorry. Plus, It would be one race to 5, 1 days to six. So the final answer would be three days to 6 -3. 7. How many onto functions are there from a set with?Summary: The formula to find the number of onto functions from set A with m elements to set B with n elements is nm - nC1(n - 1)m + nC2(n - 2)m - ... or [summation from k = 0 to k = n of { (-1)k . How do you find the number of onto a function?Number of Surjective Functions (Onto Functions) If a set A has m elements and set B has n elements, then the number of onto functions from A to B = nm – nC1(n-1)m + nC2(n-2)m – nC3(n-3)m+…. - nCn-1 (1)m. How many onto functions are there from a set of six elements to a set of three?Uh From six element to a set of three elements would be given by three days to six three. to raise to six mile plus three C 2 to raise two, sorry. Plus, It would be one race to 5, 1 days to six. So the final answer would be three days to 6 -3.
How do you find how many onto functions there are?Number of Surjective Functions (Onto Functions)
If a set A has m elements and set B has n elements, then the number of onto functions from A to B = nm – nC1(n-1)m + nC2(n-2)m – nC3(n-3)m+…. - nCn-1 (1)m. Note that this formula is used only if m is greater than or equal to n.
How many different onto functions are there from set with 5 elements to sets with 3 element?Image of each element of A can be taken in 3 ways. ∴ Number of functions from A to B = 35 = 243.
How many onto functions are there from an N element?Therefore total number of onto functions is 2n – 2.
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