Questions tagged [binomial-coefficients]
Coefficients involved in the Binomial Theorem. $\binom{n}{k}$ counts the subsets of size $k$ of a set of size $n$.
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questions
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How do you find the coefficient of $x$ in $(x + 1)^2$?
I want to learn how can I find out the coefficient of the variable $x$ in the expression $(x + 1)^2$. It is a case of a perfect square expansion.
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1answer
25 views
Sum of the product of binomial coefficients
I am trying to prove $\sum_{j=1}^{n-k}{n \choose j}{n-k-1 \choose j-1}={2n-k-1 \choose n-k}$. I tried to apply Vandermonde's Identity, however I have not been able to.
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How was this identity found: $\sum_{k=1}^{n}\binom{n}{k}\sum_{1\le i\le j\le k}\frac{1}{ij}=\sum_{1\le i\le j\le n}\frac{2^n-2^{n-i}}{ij}$?
For any postive integer $n$, prove
$$\sum_{k=1}^{n}\binom{n}{k}\sum_{1\le i\le j\le k}\dfrac{1}{ij}=\sum_{1\le i\le j\le n}\dfrac{2^n-2^{n-i}}{ij}.$$
This problem is 2019 AMM problem. I don't ask ...
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2answers
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Show $\sum_{k=1}^n {k+1\choose 2}{2n+1\choose n+k+1}={n\choose 1}4^{n-1}$
I've been attempting to show that:
$$\sum_{k=1}^n {k+1\choose 2}{2n+1\choose n+k+1}={n\choose 1}4^{n-1}\\
\sum_{k=2}^n {k+2\choose 4}{2n+1\choose n+k+1}={n\choose 2}4^{n-2}$$
Can anyone give some ...
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1answer
38 views
How to interchange the two summations and simplify?
How can we simplify the following summation?
$$\sum_{i=0}^{r-j}\sum_{l=0}^{i+j-1}(-1)^{i-1}\binom{i+j}{i}\binom{r-i-1}{r-j-i}F(r,j,l).$$
Here, $F(r,j,l)$ is independent of $i$. So, how are the limits ...
2
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1answer
37 views
Proove that Abel polynomials are of binomial type
There are many places where the statement that these polynomials are of binomial type but no one actually proves that
Let
$$
P_n(x) = x(x + an)^{n-1}
$$
Prove that
$$
P_n(x+y) = \sum_{k=0}^{n}C_n^{k}...
2
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2answers
38 views
Sum of product of NB combinations
While trying to solve the PMF of two independent NB random variables, I end up with a summation of the product of two combinations:
$$\sum_{j=0}^{k} {j+r-1 \choose j} {k-j+s-1 \choose k-j} $$
...
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0answers
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How do i evalute the following binomial series [closed]
How do I show that $$\sum_{r=0}^{n}{2n\choose 2r}{n-r\choose j}={n\choose j}\frac{\sqrt \pi}{(n-1)!}\frac{(2n-j-1)!}{(n-j-\frac{1}{2})!}$$ where $j\leq n$.
Any help will be appreciated.
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1answer
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Interpolation recurrence relation limit
I'm dealing with a problem (background at the end) where I'm using linear interpolation. I'm trying to figure out number of steps required to get within a specified limit, and the interpolation factor ...
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1answer
51 views
Find the generating function and the number of integers solutions for $x_1 + x_2 + x_3 + x_4 = r$, where $-3 \leq x_i \leq 3$. [closed]
Well, we have $x_1 + x_2 + x_3 + x_4 = r$ where $x_i\in\{-3,-2,-1,0,1,2,3\}$
Then the generating function is given by $f(x)=(x^{-3}+x^{-2}+x^{-1}+x^{0}+x^1+x^2+x^3)^4 = \frac{(x^{6}+x^{5}+x^{4}+x^{3}+...
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3answers
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Closed for solution for $\sum_{k = 0}^{n} Q^{k} ( 1 - Q) ^ {k}$
I know the binomial expansion formula:
$$
(1 + x)^n = \sum_{k = 0}^{n} {n \choose k}x^k
$$
However, I am trying to find (if there is any) a closed-form solution for the following equation.
$$
\sum_{...
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1answer
50 views
Count the number of permutations in a decreasing space
Let me start off by saying that I'm not a mathematician, so this is probably an easy problem to solve, but I haven't been able to yet..
The problem is that I want to place $n$ objects on a grid with $...
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1answer
18 views
Calculating binomial coefficient
I am familiar with:
$$
\binom{n}{k} = \frac{n!}{k!(n-k)!}
$$
Now, i have an excercise in calculus which gets to the point that we should calculate:
$$
\frac{\binom{3n+1}{n+1}}{\binom{3n}{n}}
$$
...
1
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1answer
35 views
Radius of convergence for binomial series (2)
I'm having trouble calculating the radius of convergence for for the following binomial series.
More in detail, I'm having trouble finding $c_k$ and $c_{k+1}$ for the following series:
$$ \sum_{k=0}^...
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0answers
22 views
Understanding a summation involving “$n$ choose $k$” that starts from $k=0$ and then $k$ goes negative?
Algorithm 3 in Peter Borwein's paper (2000) offers this :-
$$\zeta (s)=\Biggl(\frac{1}{2^n(1 - 2^{1 - s})}\sum_{j=0}^{2n-1}\frac{e_j}{(j+1)^s}\Biggr)+\gamma_{n}(s)$$
Where
$$e_j=(-1)^j\Biggl(\sum_{k=0}...
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1answer
43 views
How To Determine If $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left(\sum_{k=0}^{n-1}\binom{2k}{k}\binom{k}{n-k}\right)$ Converges or Diverges?
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left(\sum_{k=0}^{n-1}\binom{2k}{k}\binom{k}{n-k}\right)$$
Question : How do i determine if the above Series Converges to Diverges?
I have no idea where to begin ...
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0answers
35 views
Consider a branching process $\{X_n , n \geq 0 \}$ in which the offspring distribution is binomial $(k,p)$
Consider a branching process $\{ X_n , n \geq 0 \}$ in which the offspring distribution is binomial $(k,p)$. Find probability of ultimate extinction when $k = 3$.
So I've tried this:
$P_k = P(\...
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0answers
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Minimum n in order to have at least C combinations of order r?
I am a bit stuck on this problem. How do I find the minimum number of elements n which ensure that I have at least C combinations of order r ? Basically, for a given C and r, find the minimum n which ...
0
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1answer
37 views
identity for sum of binomial coefficients
I am trying to prove that $$\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n}{2k+1}(-1)^k = \sqrt{2^n}\sin{}n\pi/4$$
This follows from the subtraction of $\sum_{k=0}^{...}\binom{n}{4k+1}$ and $\sum_{...
2
votes
1answer
34 views
Is the following partial binomial sum with vanishing parameter converging to zero?
Suppose that
$(k_m)_{m\in \mathbb{N}}$ is a sequence of natural numbers such that $k_m \to\infty$ and $\frac{k_m}{m} \to0$ as $m \to \infty$;
$(p_m)_{m \in \mathbb{N}} \subset(0,1)$ is such that $...
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0answers
34 views
Proof of summation converging to 1 [duplicate]
Can someone prove this?$$\sum_{k=n}^{2n} 2^{-k}\dbinom{k}{n}=1$$
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votes
2answers
87 views
Represent $\sum_{k=1}^{n} k^{2}$ in terms of binomial coefficient.
Came across a probability problem that is sort of challenging for a beginner in a sense that I may have not seen or came across a lot of binomial identities.
What I am looking for is to see if there ...
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0answers
20 views
combination problem with addition [closed]
Please tell me what are the ways getting 37 by adding 6 digit using 1 to 17.No restrictions.
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How would I represent combinations of fixed subsets?
Given a friendship group of $n$ people, the binomial coefficient ${n \choose k}$ can tell me how many combinations of $k$ friends there are who could meet, e.g., if there are $10$ people and we meet ...
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Solving $1+2+3+4+ \cdots+ n=$n(n+1)/2)$ $through use of binomial coefficient
I am trying to understand proofs of the $1+2+3+4+ \cdots$ series.
I'm puzzled by the 3rd point of this post where it is solved by binomial coefficient
https://math.stackexchange.com/a/2288/777575
...
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2answers
24 views
Calculating the number of poker hands that have exactly 4 different denominations
I'm been working through some examples of questions mentioned in the title and came across this one that I wasn't entirely if I was approaching this correctly.
My initial approach to solving this ...
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votes
2answers
73 views
Combinatorial identity: $\sum\limits_{k=0}^{i\land j}\binom ik(-1)^k\binom{i+j-k}i=1$
Let $i,j\in\mathbb Z_{\ge0}$ be nonnegative integers. How can we prove
$$\sum_{k=0}^{i\land j}\binom ik(-1)^k\binom{i+j-k}i=1?$$
(Here, $i\land j=\min(i,j)=\min\{i,j\}=\min(\{i,j\})$ is the minimum of ...
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4answers
75 views
Number of ordered Pairs satisfying $4^m-3^n=1$
Find the Number of ordered Pairs $(m,n)$ of positive integers satisfying $4^m-3^n=1$
Mt try:
Trivially $m=n=1$ satisfies
Let $m \gt 1$
$$4^m-3^n=(1+3)^m-3^n=1$$
$\implies$
$$3\binom{m}{1}+3^2\binom{...
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1answer
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Combinatorial identity: $\sum_{i=0}^{k}\binom{n}{i}p^{i}q^{n-i}+ \sum_{i=k}^{n-1}\binom{i}{k}p^{k+1}q^{i-k}=1$.
While answering a recent question I came across an unexpected identity:
$$
\sum_{i=0}^{k}\binom{n}{i}p^{i}q^{n-i}+
\sum_{i=k}^{n-1}\binom{i}{k}p^{k+1}q^{i-k}=1.\tag1
$$
valid provided that $p+q=1$.
...
1
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1answer
33 views
Combinatorial problem: black and white balls divided into k groups with limits, and next picking a sequence of only black balls.
I have the following combinatorial problem: Let's assume we have $N$ balls: $W$ white, and $B$ black balls ($N = B + W$).
Step 1: The first thing we do is to randomly divide these $N$ balls into $K$ ...
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0answers
78 views
Reasoning about least common multiples and their relationship to binomial coefficients
I have been fascinated by this inequality which I believe is valid:
Let:
$x \ge n$ be integers
lcm$(a,b,c)$ be the least common multiple of integers $a, b, c$
$$\text{lcm}(x+1, x+2, \dots, x+n) \...
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1answer
50 views
Evaluation of a tricky binomial sum
The Question:
To prove that:
$\frac{3!}{2(n+3)} = \sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}$
My Attempt:
I start off by writing $\sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}...
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votes
2answers
94 views
prove: $\displaystyle\sum_{k=0}^{n}\frac{1}{k+1}\binom{2k}{k}\binom{2n-2k}{n-k} = \binom{2n+1}{n}$
prove the following identity:
$\displaystyle\sum_{k=0}^{n}\frac{1}{k+1}\binom{2k}{k}\binom{2n-2k}{n-k} = \binom{2n+1}{n}$
what I tried:
I figured that: $\displaystyle\binom{2n+1}{n} = (2n+1) C_n$
...
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1answer
32 views
Solving $C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$.
$C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$
I did find by brute force the solutions $n=1$ and $n=4$, through the inequalities $2x \le 13, x \ge 0 \implies x \in \{0,1,2,3,4,5,6\}$
But is there a more ...
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0answers
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Is there any way to generate solutions to a linear system given the domain of the terms (and some binomial products)?
Consider
$2l + 3m +4n =11$
I have constraint
$$ 0 \leq l \leq 4$$
$$ 0 \leq m \leq 4$$
and,
$ 0 \leq n \leq 4$
Given this constraint would there be any systematic way to write do all the ...
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2answers
51 views
Summation of binomial-like terms
Simplification of the two expressions
$$S_1=n\sum_{k=0}^{n-1} {n-1 \choose k} \frac{(-1)^k}{k+1}$$
and
$$S_2=n\sum_{k=0}^{n-1} {n-1 \choose k} \frac{(-1)^k}{k+3}$$
It seems that $S_1=1$, for example; ...
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0answers
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Can anyone help me to understand how this upper and lower bound for the binomial coefficient are derived?
I found this very interesting upper and lower bound for the binomial coefficient that is taken from a book that I have not heard of.
Could someone help me understand how it is derived?
Here is the ...
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1answer
45 views
$\mathbb{P}(X>Y)$ for $X,Y$ two Poisson
Two teams have to play a final of a tournament. Team A score a number of goals that can be shaped like a random variable $X \sim Poi(\lambda_{A}=2.5)$. Team B score a number of goals that can be ...
2
votes
2answers
72 views
Having $\binom{\lambda}{n}=n+1$, which is $\lambda$?
One question, if I've got this, $\binom{\lambda}{n}=n+1$, which has to be the value of $\lambda$?
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0answers
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Prove or disprove that $\frac{(x+n)!}{x!}$ is not divisible by $n$ distinct primes where each prime is greater than $\frac{(x+n)e}{n}$
Is it true that for integers $n > e, x \ge n$, it is impossible for $\dfrac{(x+n)!}{x!}$ to be divisible by $n$ distinct primes where each prime is greater than $\dfrac{(x+n)e}{n}$?
Example:
$x=n=...
8
votes
2answers
208 views
Is there a closed form for $\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x$?
Do we know if there is a closed form for
$$
I :=\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x\mathrm{?}
$$
Wolfram alpha gives an approximation of $2.66989$ which may be equivalent to:
$$10\...
1
vote
2answers
48 views
Finding an identity to simplify this combinatorics solution
Steve flips 499 coins and Marissa flips 500 coins. What is the probability that Marissa flips more heads than Steve does?
We use casework for each of the possible number of heads that Steve flips. ...
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4answers
35 views
Floating Numbers in Combinations
What could be the answer to
${\displaystyle {\binom {2.5}{2}}}$
is it defined or considered as $0$ or $1$?
0
votes
1answer
29 views
Question about how the combination formula works
We need to make 2 different groups containing 2 each from 4 people {x,y,c,z}
(x,y) and (c,z)
(x,z) and (y,c)
(x,c) and (y,z)
so we have 3 options
but when we use the formula C(4 2), we get 6. ...
6
votes
0answers
198 views
Are these well known properties of binomial coefficients?
I apologize for the number of definitions. I did not know how to state these ideas any simpler. If anyone can help me simplify the definitions, I will be glad to shorten the details.
Let:
$x,n$ ...
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0answers
39 views
Prove combinatorially: $\sum_{r=0}^n (-1)^r \binom{n}{r}CC^{k-r}_n = 0$
$n > k > 0$. $n,k \in \mathbb{N}$
Prove combinatorially (without algebric manipulation): $\sum_{r=0}^n (-1)^r \binom{n}{r}CC^{k-r}_n = 0$
Note: $CC^{k}_n$ is the number of options to ...
2
votes
0answers
26 views
Binomial coefficients hierarchy extended to real numbers
I am working with the binomial coefficient but I'm a little stuck when it involves real numbers instead of only naturals.
I know that if $0\leq k<k'<\frac{a}{2}$ and $k,k'\in \mathbb{N}_0$, ...
0
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0answers
15 views
Find the misterious function $F(r,a,q)$.
I've the pattern bellow where we can spot the binomial coefficients. Observing the pattern, for each row, it seems that it arises from
$$
\left(\left(r a^{q} \right)^{\frac{1}{q}} +F(r,a,q)\right)^{n}
...
0
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1answer
56 views
Show that $\binom{n-1}{r-1}=\sum^{r-1}_{k=0}(-1)^k\binom rk\binom{n+r-k-1}{r-k-1}$
This is an identity that I couldn't prove.
$$\binom{n-1}{r-1}=\sum^{r-1}_{k=0}(-1)^k\binom rk\binom{n+r-k-1}{r-k-1}.$$
I believe that the identity is provable through combinatorics, but I can't ...
2
votes
1answer
95 views
Prove or disprove $\frac{(x+n)!}{(x!)\text{lcm}(x+1, \dots, x+n)} < (n-1)!$
Is it correct that for any positive integers $x,n$, that $\frac{(x+n)!}{(x!)\text{lcm}(x+1, \dots, x+n)} < (n-1)!$ where lcm is the least common multiple.
I ask because I find this relationship ...
