Number of one-one and onto functions

Question

I got this question from one of my friends:

Let $X$ be a set with exactly $5$ elements and $Y$ be a set with exactly $7$ elements. If $α$ is the number of one-one functions from $X$ to $Y$ and $β$ is the number of onto functions from $Y$ to $X$, then the value of $\frac{(β-α)}{5!}$ is ___

I get it that the value of $α$ will be $^7\mathit C _5\cdot5!$, Because number of one-one functions is just the total no of functions from $A$ to $B$ in which every $x\in X$ has a single image in $Y$. How do I calculate $β$?

[Forgive the grammatical errors, if any, English is not my first Language]

@HariRamakrishnanSudhakar Yes, with respect to sets, I am aware of the principle of inclusion and exclusion. — Doodoo28, Aug 09 '20 at 08:18
https://math.stackexchange.com/questions/334420/number-of-onto-functions see if this answers your question — Albus Dumbledore, Aug 09 '20 at 08:20
@HariRamakrishnanSudhakar "Unfortunately, a function whose range misses two members of B gets subtracted twice in that computation, and it should be subtracted only once. Thus, we have to add back in the functions whose ranges miss at least two points of B", didn't quit get this point! — Doodoo28, Aug 09 '20 at 09:00
If you do not understand the Inclusion-Exclusion Principle argument, another option would be to note that either one element of $X$ is the image of three elements of $Y$, with each of the other elements of $X$ being the image of a single element of $Y$, or two elements of $X$ are each the image of two elements of $Y$, with each of the other elements of $X$ being the image of a single of element of $Y$. — N. F. Taussig, Aug 09 '20 at 09:22

Zuy · Answer 1 · 2020-08-09T10:32:56.950

I believe that you are looking for the following

Proposition:

Let $n\geq m>0$ be natural numbers. Then the number of surjections from $\{1,\dots,n\}$ to $\{1,\dots,m\}$ equals $$\sum_{k=0}^m(-1)^k {m\choose k} (m-k)^n.$$

Proof: For $1\leq i \leq m$, let $B_i:=\left\{f:[n]\to [m]\backslash\{i\}\right\}$, where $[N]:=\{1,\dots,N\}$.

Then the number $S$ of surjections from $[n]$ to $[m]$ is given by the cardinality of $$\left(B_1 \cup \dots \cup B_m\right)^C,$$ i.e. the complement of $B_1 \cup \dots \cup B_m$ in the set of all maps from $[n]$ to $[m]$.

Hence $$S=\#\left(B_1 \cup \dots \cup B_m\right)^C=m^n-\#\left(B_1 \cup \dots \cup B_m\right),$$ where we used that the number of maps from $[n]$ to $[m]$ equals $m^n$.

Next, using the principal of inclusion-exclusion, we can replace $\#\left(B_1 \cup \dots \cup B_m\right)$ by

$$\sum_{k=1}^m(-1)^{k-1}\sum_{1\leq i_1<i_2<\dots<i_k\leq m}\#\bigcap_{l=1}^k B_{i_l}=-\sum_{k=1}^m(-1)^k\sum_{1\leq i_1<i_2<\dots<i_k\leq m}(m-k)^n.$$

This leaves us with $$S=m^n+\sum_{k=1}^m(-1)^k\sum_{1\leq i_1<i_2<\dots<i_k\leq m}(m-k)^n.$$

Lastly, $m^n$ can be taken into the first sum, and the second sum just adds up $m\choose k$ times the term $(m-k)^n$, hence the simplification $$S=\sum_{k=0}^m(-1)^k {m\choose k} (m-k)^n.$$

For you, it suffices to replace $n$ by $7$ and $m$ by $5$ in the Proposition's formula.

Number of one-one and onto functions

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