How to algebraically prove that $$n!=\sum\limits_{r=0}^n (-1)^r \binom{n}{r} (n-r)^n$$  I was trying to find number of onto functions from $A$ to $A$ containing $n$ elements.  Using the inclusion-exclusion principle I am getting  $$\sum\limits_{r=0}^n (-1)^r \binom{n}{r} (n-r)^n.$$  We can also do it by simple combinatorics, as every element has to have a pre-image and number of elements in the domain equal to the number of elements in the codomain, the number of functions is $n!.$   Is there an algebraic way to prove these two are equal?
123943
$$n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$$  Could anyone give the proof of the above equation? Thanks in advance!
19992
$p \in \mathbb{R}[x]$ and $ \forall x\in\mathbb{R} \ \ p(x) &gt; 0$. Prove that $p = \sum_{k=1}^{n}p^2_k, \ \ p_k \in \mathbb{R}[x]$. I have noticed that $deg(p)$ must be even, because $p(x)$ doesn't intersect $\{x=0\}$.  How can I proceed with the proof?
43188
I originally stumbled upon this beast whilst trying to find the number of surjective functions from $[n]=\{ 1,...,n\}$ to $[k]$. Here is my idea, denote this number $a(n,k)$ hence $a(n,k) =k^n- \#\{{f:[n]\to [k]\; | \; f \; isn't \; surjective}\}$. For every $1\leq m\leq k-1$ we have ${m\choose k}$ ways of choosing a set of $m$ elements from $[k]$ and $a(n,m)$ functions in $\{{f:[n]\to [k]\; | \; f \; isn't \; surjective}\}$ with $|im(f)|=m$ and so: $a(n,k+1)=(k+1)^n-\sum_{m=1}^k {k+1\choose m}a(n,m)$ From here I tried to use induction to get $a(n,k+1)$ to the form of a polynomial in $a(n,1)$ which is clearly $1$, but had no luck as to finding the polynomial itself. Solutions to the original problem or to the recurrence relation would by much appreciated. EDIT: from the inclusion-exclusion prinicple I got the follwing expression for $a(n,k)$ which is: $a(n,k) = \sum_{m=0}^k (-1)^m{k\choose m}(k-m)^n$ and plugging it into $a(n,k+1)=(k+1)^n-\sum_{m=1}^k {k+1\choose m}a(n,m)$ yeilds: $a(n,k+1)=(k+1)^n-\sum_{m=1}^k{\sum_{i=0}^m {k+1 \choose k+1-m,i,m-i}(m-i)^n}$ which, to be honest, looks terrible. My original question still stands, what is the closed-form expression (if there even is one) for this monster double sum?
105551
Suppose that sequences of real numbers satisfy:  \begin{align*}   a_n &amp;= \sum\limits_{r=0}^n \binom{n}{r} b_r \\ \end{align*}  Prove that:  \begin{align*}   (-1)^n b_n &amp;= \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s \\ \end{align*}  My work:  The $n=0$ case:  \begin{align*}   a_0 &amp;= b_0 \\   b_0 &amp;= a_0 \\ \end{align*}  Then $n=1$ case:  \begin{align*}   a_1 &amp;= \binom{1}{0} b_0 + \binom{1}{1} b_1 = b_0 + b_1 \\   -b_1 &amp;= \binom{1}{0} a_0 - \binom{1}{1} a_1 = a_0 - a_1 \\   b_1 &amp;= a_1 - a_0 = b_0 + b_1 - a_0 = b_1 \\ \end{align*}  The inductive step:  \begin{align*}   (-1)^{n+1} b_{n+1} &amp;= \sum\limits_{s=0}^{n+1} \binom{n+1}{s} (-1)^s a_s \\ \end{align*}  I'm not sure where to go from here. I tried substituting $a_s$ into that last equation and it didn't help. Any ideas?
56095
If $k, m, n$, are natural numbers and $k \leq n$ What is:  $$\sum_{r=0}^{m}\frac{k\binom{m}{r}\binom{n}{k}}{(r+k)\binom{m+n}{r+k}}$$
64945
I'm really iffy on combinatorial proofs in general and now that there is a sum, it's just confused me even more. Can someone try and walk me through this proof?  $$ \binom{m + n}{r} = \sum_{k=0}^r \binom{m}{k}\binom{n}{r - k} $$
68476
Here's how it is.  So I was studying some notes on Discrete Structures then I discovered this. Prove that $2^N = \binom{N}{0} + \binom{N}{1} + \binom{N}{2} + \dots + \binom{N}{N}$ Since I was new to this, I didn't quite understand it.  I tried using the binomial expansion but it confused me a little.  Can someone please help me solve this.  That would be great.
9697
I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more general way.  The left side of an identity occurs while solving another problem (concerning binomial theorem) so I am more interested in deriving the right side from the left side, else I have to remember it now onward.  EDIT: I am more interested in an algebraic proof rather than combinatorial argument or something involving calculus (however I liked  and  solution), hence I am removing the combinatorics tag.
45758
I'm studying for a midterm and need some help with proving summation     $$\sum\limits_{k=0}^n\binom{n}{k}2^k = 3^n$$   using the binomial theorem.  This is what I've been thinking so far:   In the binomial theorem, we set $x = 0$ and $y = 2$, so:  $3^n = (x+y)^n = \sum\limits_{k=0}^n\binom{n}{k}y^k$  $ = \sum\limits_{k=0}^n\binom{n}{k}2^k$  Am I getting this correct so far or completely wrong? Any help would be appreciated. Thanks.
112406
Show that every positive integer is a sum of one or more numbers of the form $2^r3^s,$ where $r$ and $s$ are nonnegative integers and no summand divides another.   From:   I see they mean that:  $$n-1 = \sum_{k=1}^{x} 2^{r_k} 3^{s_k}$$  By induction hypothesis. Then:  $$n = \sum_{k=1}^{x} 2^{r_k} 3^{s_k} + 1$$  Then how to proceed by $n/2$? How can you say that:  $$\frac{n}{2} = \sum_{k=1}^{x} 2^{r_k} 3^{s_k} $$  Where did the $1$ go? And how can you assume it works for $\frac{n}{2}$ (the hypothesis)?
16105
How do you prove that $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$? I tried to identify the sum as a binomial series, but the $4$ and the $-1/2$ puzzle me. (This series arises in studying the first passage time of a simple random walk.)
86807
Prove the identity $\sum_{k=0}^{n}\sum_{r=0}^{k} \binom{k}{r} \binom{n}{k} = 3^n$. I believe I need to use the binomial theorem here, but I don't know how to deal with the double summations.
43096
Find a closed form expression for $$\sum_{r=0}^n \dfrac{(-1)^r}{\dbinom{n}{r}}$$ where $n$ is an even positive integer.  I tried using binomial identities, but since the binomial coefficient is in the denominator, I can't do much. Also, I'm seeking a non-calculus, elementary solution, if possible. Thanks.
51836
this is a question from a Temple prelim exam, and i'm trapped in it! We have $p_n(z)=\sum_{k=0}^n\frac{z^k}{k!}$ and we have to prove that $\forall r&gt;0 \quad \exists N\in\mathbb{N}$ s.t. $p_n(z)$ has no zeros in $B_r(0)$ for $n&gt;N$.  I tried to use Rouche's theorem observing that $1+\frac{z^n}{n!}$ has the nth roots of $n!$ as complex zeros, which are outside a fixed ball for n sufficiently large, but i'm having trouble proving that $|1+\frac{z^n}{n!}|&gt;|z+\cdots+\frac{z^{n-1}}{(n-1)!}|$ in $|z|=r$ (if this is true). If anyone could help..
54052
I would appreciate if somebody could help me with the following problem  Q: Seeking a combinatorial proof $(\binom{n}{k}=\frac{n!}{k! (n-k)!} )$  $$\sum _{k=0}^n (n-2 k)^2 \binom{n}{k}=n\times  2^n$$
94106
Let $N$ be a positive integer. Then, we have  $\sum\limits_{j=1}^{N} \binom{j}{6} = \binom{N+1}{7}$.  Could anyone explain this equation a little bit? I wrote out the left hand side as $\binom{1}{6} + \dots + \binom{N}{6}$, but it did not help at much.  UPDATE  From lecture notes, I found a proposition says:     $\sum\limits_{0\leq k\leq n} \binom{k}{m} = \binom{n+1}{m+1}$.   Could anyone give a hint on how to prove it?
36137
So I came across Pascal's identity: Prove that for any fixed $r\geq 1$, and all $n\geq r$, $$ \binom{n+1}{r}=\binom{n}{r}+\binom{n}{r-1}. $$  I know you can use basic algebra or even an inductive proof to prove this identity, but that seems really cumbersome. I was wondering if anyone had a "cleaner" or more elegant way of proving it. For example, I think the following would be a decent combinatorial proof.   Proof: Let $S$ be a set with $n+1$ elements, and consider some fixed $x\in S$. There are $\binom{n+1}{r}\;\; r$-subsets of $S$--count them according to whether or not they contain $x$: there are $\binom{n}{r}$ not containing $x$, (each formed by choosing $r$ of the remaining $n$ elements in $S\setminus\{x\}$), and there are $\binom{n}{r-1}\;\; r$-sets containing $x$, (each formed by selecting an additional $r-1$ elements in $S\setminus\{x\}$).   Is that right? Are there any other efficient ways of doing it?