All Questions
51
questions
1
vote
1
answer
101
views
Demonstrating a Binomial Identity? #2 (The exclusion)
$$
\sum_{m=1}^{\lfloor j/(k+1) \rfloor}(-1)^m\binom{n}m\binom{j-m(k+1)+n-1}{n-1}
= \sum_{m=1}^{j-k} {j-k-1 \choose m-1}{n \choose m}m
$$
(Actually $\not =$ see edit end of post)
Is there a simple ...
- 526
1
vote
2
answers
162
views
Binomial identity?
$${n+k-1 \choose k}=\sum_{m=1}^{min(k,n)}{k-1 \choose m-1}{n \choose m} $$
Is there a simple way to demonstrate this equality?
Context
These are two ways of expressing the $x^k$ coefficients in $(1+x+...
- 526
0
votes
0
answers
62
views
Combinatorial interpretation of the multinomial coefficient as a product of binomial coefficients
$$
\begin{align}& \binom{n}{k_1,k_2,\dots,k_m}\\&=\frac{n!}{k_1!k_2!\cdots k_m!}\\&=\binom{k_1}{k_1} \binom{k_1+k_2}{k_2}\cdots \binom {k_1+k_2+\cdots+k_{m}}{k_{m}}\end{align}
$$
Is there ...
- 137
0
votes
0
answers
29
views
Weighted sum of specific multinomial coefficients
Let $A$ and $b$ be nonnegative integers and consider the sums
$$\sum\limits_{c=0}^{b/2}\frac{1}{4^c}\binom{A}{c,b-2c,A-b+c}$$
and
$$\sum\limits_{c=0}^{b/2}\frac{c}{4^c}\binom{A}{c,b-2c,A-b+c}.$$
I ...
- 1,147
2
votes
1
answer
116
views
Central Binomial Coefficients and Multinomial Coefficients
Premise
I was looking at the multinomial coefficients when selecting by a specific rule. Then analyzing the sum.
Given the multinonial theorem ($n > 0$):
$$
(x_1+\ldots+x_n)^n = \sum_{k_1+\ldots+...
- 177
2
votes
4
answers
258
views
Coefficient of $x^k$ in polynomial
Let $k, n, m \in \mathbb{N}, k \le n.$ Find the formula for coefficient of $x^k$ in $(x^n + x^{(n-1)} + ... + x^2 + x + 1)^m$.
answer is in this question: faster-way-to-find-coefficient-of-xn-in-1-x-...
- 311
0
votes
1
answer
76
views
is there a general combinatorics formula like n choose k but instead we choose k,l,m ect.
Say for example we have a box with $n$ balls of which $r$ are red, $b$ are blue, $g$ are green, $y$ are yellow (so $n=r+b+g+y$)
Now we draw out all of them, one after the other, without ever placing ...
- 3
1
vote
0
answers
110
views
Sum of multinomial coefficient
I want to show that $$\sum_{j,k,j+k\leq n} 3^{-n}\frac{n!}{j!k!(n-k-j)!}=1.$$
I tried to do it by induction, but this clearly seems the wrong approach because the denominator becomes very hard to deal ...
- 1,952
3
votes
4
answers
1k
views
Why does the multinomial coefficient count permutations but the binomial coefficient count combinations?
The binomial coefficient $n \choose k$ counts the number of ways to choose $k$ objects from a set of $n$ objects (order does not matter).
The more general multinomial coefficient $n \choose {n_1,n_2,.....
- 1,611
1
vote
2
answers
106
views
A certain sum of multinomial coefficients
I would like to know if there is a nice expression for the sum
$$
S(n)=\sum_{i+j=n}\binom{3i}{i,i,i}\binom{3j}{j,j,j}
$$
where $n$ is a non-negative integer. I have entered in the first few values of ...
- 35
1
vote
1
answer
158
views
I have the sequence: $1, 4, 10, 16, 19, 16, 10, 4, 1$. What is the formula to get the $i^{th}$ term of the sequence for $i=1,2,\dots,9$
I've been trying to derive this formula for quite some time now with little progress.
I've seen concepts such as Pascal's pyramids and Pascal's simplices mentioned throughout my research; however, I ...
- 11
1
vote
0
answers
73
views
Sum of products of binomial coefficient
The multinomial theorem states that
\begin{align}
(x_{1} + \dots + x_{m})^{n} = \sum_{k_1 + \dots + k_{m} = n} \binom{n}{k_1, k_2, \dots, k_{m}} x_{1}^{k_{1}} \cdots x_{m}^{k_{m}} = \sum_{k_1 + \dots +...
- 515
2
votes
1
answer
863
views
Interpretation of multinomial coefficients in terms of choosing elements from a set?
The binomial coefficients represent the coefficients on the terms in the expansion of $(x+y)^n$, but they can also be interpreted as choosing a subset of items from a set while disregarding the order ...
- 167
0
votes
0
answers
155
views
Product of Binomial coeficients
Suppose I have a set $\{n_1, n_2, \cdots, n_m\}$ such that $n=\sum\limits_{i=1}^m n_i$. Is there a way to simplify the expression
$$
\prod\limits_{i=1}^m {n \choose n_i}
$$
I feel like this might be ...
- 83
4
votes
1
answer
72
views
Simplify $\sum_{k = 0}^n\sum_{k_1+\ldots+k_{m}=n-k} \binom{nm}{k_1,\ldots,k_m,n-k_1,\ldots,n-k_m}$
Let $n$ and $m$ be positive integers. I want to find a formula for the following expression:
$$\sum_{k = 0}^n\quad \sum_{k_1+...+k_m=n-k} \quad \binom{n-k}{k_1,\ldots,k_m}\binom{nm}{n-k_1,\ldots,n-k_m,...
- 1,093