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$1.$ Prove $\forall n\in\mathbb{N}\quad n\geq5$$$ n^4-\binom{n}{1}(n-1)^4+\binom{n}{2}(n-2)^4-\binom{n}{3}(n-3)^4+\dots +(-1)^{n-2}\binom{n}{n-2}\cdot2^4+(-1)^{n-1}\binom{n}{n-1}\cdot1^4=0 $$$2.$ Prove $\forall n\in\mathbb{N}\quad$$$ n!=\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}(n-k)^n $$

Progress:
$1.$ My intuition here is that $n^4$ represent the number of all functions, say $f:A\to B$ where $|A|=4$ and $|B|=n$, though I'm not quite sure what the other elements represent according to the inclusion-exclusion principle.

$2.$ My intuition here is that the number of injective functions, say $f:A\to A$ where $|A|=n$ is equal to the number of surjective functions. The first element of the sum, $n^n$ is the total amount of functions $f:A\to A$ and apparently we define $A_{i}=\{f:A\to A|\quad\forall a\in A\quad f(a)\neq i\}$ and use the inclusion-exclusion principle.

Corrections/suggestions would be much appreciated

Slavik Egorov
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1 Answers1

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Actually the comments seem to have been helpful enough. I can write the answer by adding the details so that the problem does not seem unsolvable.

$n, m$ are positive integers and $A, B$ are finite sets that $|A|=m$, $|B|=n$. If we use the inclusion-exclusion principle, the number of surjective (onto) functions from set $A$ to set $B$ is $$ \color{blue}{n^m - \dbinom{n}{1}(n-1)^{m} + \dbinom{n}{2}(n-2)^{m} - \cdots + (-1)^{n-1}\dbinom{n}{n-1}1^{m} \\ = \sum_{r=0}^{n}(-1)^r\dbinom{n}{r}(n-r)^m} \tag{*}$$

$1.$ Let's take $n\geq5$ and $m=4$ in $(*)$. On the other hand, the number of surjective (onto) functions from set $A$ to set $B$ is $0$. So, we conclude that the first identity satisfy. $$ \color{red}{n^4-\binom{n}{1}(n-1)^4+\binom{n}{2}(n-2)^4-\binom{n}{3}(n-3)^4+\cdots +(-1)^{n-1}\binom{n}{n-1}\cdot1^4=0} $$

$2.$ Let's take $n=m$ in $(*)$. In this case, if $f:A\to B$ is surjective then $f$ is injective (one-to-one) function. So, $f$ is bijective. On the other hand, the number of bijective functions is $n!$. Hence, we get $$ \color{red}{n^n - \dbinom{n}{1}(n-1)^{n} + \dbinom{n}{2}(n-2)^{n} - \cdots + (-1)^{n-1}\dbinom{n}{n-1}1^{n} = n!} $$

Also, we can find a third identity with using $(*)$.

$3.$ For all $n$ positive integers, $$ \color{red}{\sum\limits_{k=0}^{n} (-1)^k\dbinom{n}{k}(n-k)^{n+1} = n! \dbinom{n+1}{2}}$$

Proof: Let $|A|=n+1, |B|=n$. By the inclusion-exclusion principle, the sum in the left had side gives the number of surjective functions. Now let's calculate the number of surjective functions directly. The image of the two elements in $A$ must be the same. We can select these two elements in the $\dbinom{n+1}{2}$ ways. We can choose their image in $B$ in $n$ ways. The other $n-1$ elements in $A$ will each have a different image. These are formed in $(n-1)!$ different ways. By the multiplication principle, we get the result as $\dbinom{n+1}{2}n(n-1)! = n! \dbinom{n+1}{2}$.

scarface
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