Where A = {1,2,3,4,5,6} and B = {a,b,c,d,e}. Conversely, each ordered partition of A into k non-empty subsets defines a surjection f: A B Therefore, the number of ordered partitions of A coincides with the number of surjections from A to B. Assign images without repetition to the two-element subset and the four remaining individual elements of A. I'll write the argument in a somewhat informal "physicist" style, but I think it can be made rigorous without significant effort. Let the two sets be A and B. and $$\sigma^2=\frac14 [1-e^\alpha\ln(1+e^{-\alpha})]=\frac14 [1-e^{-1.366}(1.59)]=.149$$ 2 See answers Brainly User Brainly User A={1,2,3,.....n}B={a,b}A={1,2,3,.....n}B={a,b} A has n elements B has 2 elements. = 1800. So, heuristically at least, the optimal profile comes from maximising the functional, subject to the boundary condition $f(0)=0$. But we want surjective functions. S(n,m) is bounded by n - ceil(n/3) - 1 and n - floor(n/4) + 1. $$B=\frac{re^{2r}-(r^2+r)e^r}{(e^r-1)^2}.$$, I found this paper of Temme (available here) that gives an explicit but somewhat complicated asymptotic for the Stirling number S(n,m) of the second kind, by the methods alluded to in previous answers (generating functions -> contour integral -> steepest descent), Here's the asymptotic (as copied from that paper). To look at the maximum values, define a sequence S_n = n - M_n where M_n is the m that attains maximum value for a given n - in other words, S_n is the "distance from the right edge" for the maximum value. There are 3 ways of choosing each of the 5 elements = [math]3^5[/math] functions. Although his argument is not as easy as the complex variable technique and does not give the full asymptotic expansion, it is of much greater generality. I should have said that my real reason for being interested in the value of m for which S(n,m) is maximized (to use the notation of this post) or m!S(n,m) is maximized (to use the more conventional notation where S(n,m) stands for a Stirling number of the second kind) is that what I care about is the rough size of the sum. since there are 4 elements left in A. Saying bijection is misleading, as one actually has to provide the inverse function. By standard combinatorics number of surjection is 2n−2. This calculation reveals more about the structure of a "typical" surjection from n elements to m elements for m free, other than that $m/n \approx 1/(2 \log 2)$; it shows that for any $0 < t < 1$, the image of the first $tn$ elements has cardinality about $f(t) n$. The sum is big enough that I think I'm probably not too concerned about a factor of n, so I was prepared to estimate the sum as lying between the maximum and n times the maximum. Among other things, this makes $x_0$ and $t_0$ bounded, and so the f(t_0) term is also bounded and not of major importance to the asymptotics. The number of surjections between the same sets is [math]k! It does seem though that the maximum is attained when $m/n = c+o(1)$ for some explicit constant $0 < c < 1$. EDIT: Actually, it's clear that the maximum is going to be obtained in the range $n/e \leq m \leq n$ asymptotically, because $m! In principle this is an exercise in the saddle point method, though one which does require a nontrivial amount of effort. Thus, for the maximal $m$ , the number of maps from $n$ to $m+1$ is approximatively 4 times the number of maps from $n$ to $m$ . If one fixes $m$ rather than lets it be free, then one has a similar description of the surjection but one needs to adjust the A parameter (it has to solve the transcendental equation $(1-e^{-A})/A = m/n$). The translation invariance of the Lagrangian gives rise to a conserved quantity; indeed, multiplying the Euler-Lagrange equation by $f'$ and integrating one gets, for some constants A, B. To learn more, see our tips on writing great answers. it is routine to work out the asymptotics, though I have not bothered to = \frac{1}{2-e^t} $$ If f : X → Y is surjective and B is a subset of Y, then f(f −1 (B)) = B. $$\Pr(\text{onto})=\frac1{m^n}m! Use MathJax to format equations. (The fact that $h$ is concave will make this maximisation problem nice and elliptic, which makes it very likely that these heuristic arguments can be made rigorous.) Every function with a right inverse is necessarily a surjection. Check Answer and Solutio I'm assuming this is known, but a search on the web just seems to lead me to the exact formula. The same sets is where denotes the Stirling numbers of the 5 elements = math! 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