5
votes
2answers
68 views

Challenge: How to prove this identity between bi- and trinomial coefficients?

This question is the continuation of its predecessor. Using the convention that trinomial coefficients $$ \binom{n}{k_1,k_2,k_3}=\frac{n!}{k_1! k_2! k_3!} $$ are zero if $k_i<0$ or $\sum_i k_i\neq ...
1
vote
2answers
39 views

Factorial as a sum. Insight appreciated

I recently posted an answer to a question about ways to express the factorial function as a sum. I posted the following formula, which I discovered several years ago and I haven't seen anywhere else: ...
3
votes
4answers
130 views

Proving Combinatorical Summation: $n!=\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n$ [duplicate]

been stuck with this question for the last few hours, any help would be appreciated. $$ {\large n! = \sum_{k = 0}^{n}\left(-1\right)^{k}{\,n\, \choose \,k\,} \left(\,n - k\,\right)^{n}} $$ what I ...
2
votes
0answers
58 views

How to prove these indentities? [closed]

How to prove these indentities? $\displaystyle \sum \limits_{k\geq0} {2n\choose 2k-1}{k-1\choose m-1}=2^{2n-2m+1}{2n-m\choose m-1}$ $\displaystyle \sum \limits_{k=0}^{m} {m\choose k}{n+k\choose ...
2
votes
1answer
50 views

Calculate sum wtih binomial coefficients

I need help with finding the sum of $\sum \limits_{k=0}^{n} \frac{1}{k+1}{n\choose k}x^{k+1}$
0
votes
2answers
45 views

How to calculate this sum

How do you calculate this sum $ \sum \limits_{k=1}^{n} \frac{k}{n^k}{n\choose k}$ ?
2
votes
1answer
71 views

Using combinatorial reasoning to show $n!=\binom{n}{0}D_n+\binom{n}{1}D_{n-1}+\dots+\binom{n}{n}D_0$

How can one use combinatorial reasoning to show that $$n!=\dbinom{n}{0}D_n+\dbinom{n}{1}D_{n-1}+\dbinom{n}{2}D_{n-2}+....+\dbinom{n}{n-1}D_1+\dbinom{n}{n}D_0$$ Now $D$ stands for deranged which is a ...
0
votes
7answers
152 views

Calculating $\displaystyle\sum_{i=1}^{n} \binom{i}{2}$

Show $\displaystyle\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$. I'm thinking right now (though not getting anywhere with it) that I want to expand out the summation portion to $i!/2!(i-2)!$ and ...
9
votes
4answers
353 views

Evaluate a finite sum with four factorials

Given positive integers $k, m, n$ such that $1 \leq k \leq m \leq n$. Evaluate $$ \sum^{n}_{i\mathop{=}0}\frac{1}{n+k+i}\cdot\frac{(m+n+i)!}{i!(n-i)!(m+i)!}$$ Any hints? I'm stuck on ...
0
votes
1answer
91 views

Closed form of $n!\sum_{k=3}^{n-1}{{n-2}\choose{k-1}}$

$n$ is given, and it takes part in the following formula. $$n!\sum_{k=3}^{n-1}{{n-2}\choose{k-1}}$$ Is there a nicer way for expressing it? Without the summation sign?
1
vote
1answer
67 views

Binomial coefficient properties

On "theoretical computer science cheat sheet" I found a special formula which is: $$ {n \choose k} = (-1)^k {k-n-1 \choose k}$$ But when I try to expand the value of ${k-n-1 \choose k}$ I have ...
3
votes
1answer
107 views

Sum of fraction of factorials

Can anybody explain this? $$\sum\limits_{k=1}^{\frac{m-1}2}\frac{(2k)!(2m-2k)!}{(2k-1)(2m-2k-1)k!^2(m-k)!^2}=\frac{(2m)!}{(2 m-1)m!^2}$$ I did actually simplify this to: ...
4
votes
1answer
85 views

Simple method for $\frac{(2n+1)!}{(n!)^{2}}$ divide $lcm(1,2,\ldots,2n+1)$

The question is to prove that $\frac{(2n+1)!}{(n!)^{2}}$ divides $lcm(1,2,\ldots,2n+1)$. This seems like it should be a simple question, but try as I might, I can't seems to find any way that does ...
5
votes
4answers
247 views

factorial as difference of powers: $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$?

The successive difference of powers of integers leads to factorial of that power. I think this is the formula $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$ But I found no proof on internet. Please ...
0
votes
1answer
90 views

Theory behind multiplication & combinations?

If with the Binomial Coefficient we try to find the possible combinations $\binom{n}{k}$ where $n$ is equal to $k$ what is the theory behind factorial resulting in the correct solution? E.g. ...
2
votes
0answers
96 views

Compute $(-1)^n\sum_{k=1}^n (-1)^k\frac{(k+n-1)!}{(k-1)!(k-1)!(n-k)!}$

Compute $(-1)^n\sum_{k=1}^n (-1)^k\frac{(k+n-1)!}{(k-1)!(k-1)!(n-k)!}$ Define $a_{k,m}=\frac{(-1)^{k+m}(n+k-1)!(n+m-1)!}{(k+m-1)[(k-1)!(m-1)!]^2(n-m)!(n-k)!}$ Compute ...
3
votes
2answers
443 views

German tank problem, simple derivation [duplicate]

I was reading the recent question on the German tank problem, and had trouble with one of the derivations in this section. $$\sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom N k} = ...
2
votes
1answer
69 views

Why is this formula for $(2m-1)!!$ correct?

Numerically calculating the sum of the squares of the $m$th row of Pascal's triangle, I found that for at least the first $10$ or so cases $$\sum_{i=0}^m \binom{m}{i}^2=\frac{(4m-2)!!!!}{m!}$$ Where ...
5
votes
3answers
277 views

Dividing factorials is always integer

Is there a simple way to show that $$n!\over r!(n-r)!$$ is always an integer?
3
votes
5answers
164 views

Can $\frac{n!}{(n-r)!r!}$ be simplified?

I'm trying to calculate in a program the number of possible unique subsets of a set of unique numbers, given the subset size, using the following formula: $\dfrac{n!}{(n-r)!r!}$ The trouble is, on ...
5
votes
4answers
546 views

Proving that ${n}\choose{k}$ $=$ ${n}\choose{n-k}$

I'm reading Lang's Undergraduate Analysis: Let ${n}\choose{k}$ denote the binomial coefficient, $${n\choose k}=\frac{n!}{k!(n-k)!}$$ where $n,k$ are integers $\geq0,0\leq k\leq n$, and ...
1
vote
4answers
556 views

Evaluate a sum involving n choose r

Evaluate: $\sum_{k=0}^6 (-1)^k \binom{6}{k}$ where $\binom{n}{r}= \frac{n!}{r!(n-r)!}$. I'm unsure how to compute the part with $\binom{6}{k}$, it should be something along the lines of ...
2
votes
3answers
122 views

Factorial Equality Problem

I'm stuck on this problem, any help would be appreciated. Find all $n \in \mathbb{Z}$ which satisfy the following equation: $${12 \choose n} = \binom{12}{n-2}$$ I have tried to put each of them ...
3
votes
2answers
67 views

Identity of binomial series with factorial.

I'm looking for a simple identity for the formula: $$ \sum_{k = 0}^{p} \binom{p}{k} \cdot k! \cdot x^k $$ In words, I have $p$ "players" who can choose to play or not (every player is represented by ...
2
votes
1answer
226 views

How to perform the summation/addition of binomial coefficients?

From my textbook: $$ \begin{align} \sum_{k=0}^n \binom {m+k}m &= \binom {m+n}m + \sum_{k=0}^{n-1} \binom {m+k}m\\\\\\ &= \binom {m+n}m + \binom {m+n}{m+1}\\\\\\ &= \binom {m+1+n}{m+1} ...
10
votes
1answer
139 views

Can this product be written so that symmetry is manifest?

Let $i,$ $j,$ $k$ be nonnegative integers such that $i+j+k$ is even. The expression $$(-1)^{j+k}\binom{i+j+k}{i,j,k}\prod_{\ell=0}^{k-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}$$ apparently computes the ...
1
vote
0answers
50 views

Approximation of factorial - Stirling formula [duplicate]

Possible Duplicate: Elementary central binomial coefficient estimates How can I prove that $$ \binom{n}{n/2} = \Theta\left(\frac{2^n}{\sqrt n}\right) $$ I tried with Stirlings ...
3
votes
0answers
135 views

Prove: $\frac{(2px)!}{((px)!)^2}\equiv\frac{(2x)!}{((x)!)^2}\pmod{p^2}$

How can I prove the following, where $p$ is a prime and $x$ a positive integer? $$\dfrac{(2px)!}{((px)!)^2}\equiv\dfrac{(2x)!}{((x)!)^2}\pmod{p^2}$$ I'm not sure if it is actually true, but I tested ...
11
votes
1answer
633 views

Factorial canceling on expansion of binomial coefficients on Concrete Mathematics

On Concrete Mathematics section 5.5, which is teaching the hypergeometric functions, generalized factorials is defined as: \[ \frac 1 {z!} = \lim_{n \to \infty} \binom{n+z}{n}n^{-z} \] where \[ ...
0
votes
2answers
130 views

What is the minimum number of friends Sally can have if she can invite a different group of friends to her house every night for a year?

The question: Sally has $N$ friends and likes to invite them over in small groups for dinner. She calculated that she can invite a different group of $3$ friends to dinner at her house every night for ...
4
votes
2answers
864 views

Factorial division using Pascal's triangle.

I want to get values of factorial divisions such as 100!/(2!5!60!)(the numbers in the denominator will all be smaller than the numerator, and the sum of the ...
21
votes
6answers
5k views

prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer

Prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer. For clarity: the denominator is the only part being squared. My thought process: The numerator is the product of the first n even ...
6
votes
3answers
1k views

Inductive proof for the Binomial Theorem for rising factorials

I want to proove the following equality containing rising factorials $$(x+y)^\overline{n}\overset{(*)}{=}\sum_{k=0}^n\binom{n}{k}x^\overline{k}y^\overline{n-k}.$$ For $n=1$ this equality is ...
9
votes
2answers
4k views

Approximating the logarithm of the binomial coefficient

We know that by using Stirling approximation: $\log n! \approx n \log n$ So how to approximate $\log {m \choose n}$?
1
vote
1answer
108 views

Combination Problem with a Variable

I have the following problem: $_xC_6$ = $_xC_4$ I expand both sides to: $$\frac{x!}{[(x-6)!]6!} = \frac{x!}{[(x-4)]!4!}$$ Next I multiply both sides by the denominator of the right-hand ...