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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Growth of binomial recurrence with different initial conditions

The binomial coefficients $\binom{n}{r}$ satisfies $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$. This means it is a solution of the equation $f(n,r)=f(n-1,r)+f(n-1,r-1)$, with initial conditions ...
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How can I prove this property of binomial coefficients?

I was playing around with binomial coefficients and binomial expansions and I came across an interesting identity: $$ \sum_{k=0}^n\frac{1}{k+1}{n \choose k}x^k=\frac{(x+1)^{n+1}-1}{(n+1)x}$$ I have ...
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1answer
74 views

Binomial Series. Product series of coefficients [closed]

How to solve this question? Please provide hints only.
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1answer
65 views

Sum of binomial coefficients $\sum_{k=n}^{r}\binom{k}{n}$

Is is possible to find the sum of a binomial coefficient series like: $\sum_{k=n}^{r}\dbinom{k}{n}$? Just a random thought.
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1answer
13 views

bounds on binomial coefficients

Do the standard upper bounds on the binomial coefficient $\binom{n}{k}$ still work well if $k=f(n)$ (by standard i mean for example $(\frac{en}{k})^{k}$ and $\frac{n^{k}}{k!}$)? In particular if ...
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2answers
29 views

Which rule is applied to define the operator precedence for factorial

Please apologize the question, I struggled with finding a good formulation in the first place: Looking at $\binom{2n}{k}$ it is very clear that for n,k integer and n>k we can solve it by calculating: ...
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2answers
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Finding a combinatorial argument for an interesting identity.

Consider the following identity $${{n}\choose{0} }{m\choose n} + {n\choose 1}{{m+1} \choose n}+ \cdots {n\choose n}{{m+n} \choose n} =\sum_{i=0}^{\min (m ,n)} {n\choose i}{m\choose i}2^i$$ The ...
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1answer
104 views

How to calculate sum of combinations with different n and k

Input: $[X,Y]$ and $L$ Output : no of increasing sequence of length L and all elements should be $X\le i \le Y$ e.g: for $[6,7]$ and $2$ sequences are $6,66,67,7,77.$ For the above question my ...
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1answer
47 views

Showing that this sum is equal to the fibonacci numbers

How do I show that the following sum is equal to the fibonacci numbers? Atleast numerical evaluation suggests it is $$ \sum_{k=0}^{\lceil n/2\rceil}\binom{n+1-k}{n+1-2k} $$ The image below shows how ...
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1answer
32 views

Calculate $\binom{n}{k}\pmod{10^6+3}$

I want to calculate the value of the following: $$\binom{n}{k}\pmod{10^6+3}$$ $10^6+3$ is prime if it may help. What is the math behind this? I can only understand basic modular arithmetic.
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1answer
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combinatorial of $\binom{10^9} {r}$ while $1 \leq r \leq10^9 \pmod {10^6+3}$

How to calculate $ \binom{n}{r} \mod m$ when $1\leq n,\: r\leq 10^9$ and $m=10^6+3$. I have tried by making Sieve of factorial and multiplicative inverse $10^6+3 \mod m$. is there any solution in ...
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0answers
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how to represent a binomial coefficient in terms of a series?

I have to find the power series for (n+m C m) or (n+m C m ) - 1 i.e representing it in some sort of power . Is it even possible ? P.S. :- Thanks in advance .
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3answers
74 views

Compact form of sum $\sum\limits_{k=0}^m (-1)^k \binom{n}{k} \binom{n}{m-k}$

How to find compact form of the sum $$\sum\limits_{k=0}^m (-1)^k \binom{n}{k} \binom{n}{m-k}$$ It looks like it's connected with Vandermonde's identity but I couldn't get to the solution.
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Summation Of Series of $\binom{x+k}{k+1}$ where $k$ is $0$ to $n$ [duplicate]

Want The formula or to find The Sum of Series where $$S=\sum_{k=0}^n \binom{x+k}{k+1}$$ where $x$ is any constant $\geq 1$ and $n$ is another constant.
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2answers
116 views

How to calculate $\sum_{m=1}^{N}\binom{m+k-1}{m}$. [closed]

What would be a simplified formula for $\displaystyle \sum_{m=1}^{N}\binom{m+k-1}{m}$ for a given number $k$ and any number $N$?
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1answer
76 views

Summation of binomial coefficients [duplicate]

Is there a closed formula for: $\sum_{i=1}^{N}{\binom{i+k}{i}}$ ( k is a constant whole number )
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1answer
26 views

$\sum^n_{r=0} (-1)^r C_r(1/2^{r}+3^r/2^{2r}+7^r/2^{3r}+\cdots\infty)$

$$\sum^n_{r=0} (-1)^r C_r(1/2^{r}+3^r/2^{2r}+7^r/2^{3r}+\cdots\infty)$$ is equal to? How to approach this problem?Hints please!!! BTW $C(r)$ stands for $(n)C(r)$
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66 views

Bonferroni Inequalities

Let $k$ and $m$ be positive integers with $k>m$. Then the partial sums of $$ 1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m} $$ has alternating signs. (The partial sums of the ...
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1answer
122 views

Complex Analysis proof of multinomial expression

I've recently come across the following identity $$ \displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n} $$ A nice complex analysis proof (by Felix Marin, here) follows as: ...
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Which term of the binomial expansion of $\left(1+\sqrt{2}\right)^{50}$ is the greatest?

Which term of the binomial expansion of $\left(1+\sqrt{2}\right)^{50}$ is the greatest? How can I find it, without comparing all 51 values? Is there a quicker way to do it? (The solution says ...
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2answers
46 views

Prove that $\frac{1}{(1-x)^k}$ is a generating function for $\binom{n-k-1}{k-1}$

On my discrete math lecture there was a fact that: $\frac{1}{(1-x)^k}$ is a generating function for $a_n=\binom{n-k-1}{k-1}$ I'm interested in combinatorial proof of this fact. Is there any simple ...
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binomial identity with negatives

Prove that $$\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}(k+1)^n=0\;.$$ I tried finding a combinatorial interpretation but to no avail. Here is a combinatorial statement, however crappy. Suppose we have $n$ ...
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Interpretation of a combinatorial identity

I am trying to find an combinatorial interpretation for the following combinatorial identity involving iterated binomial coefficients, which appeared in the November 1980 edition of The American Math ...
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3answers
39 views

Does $^nC_{n+1}$ exist? [duplicate]

is there any value assigned to $^nC_{n+1}$? My teacher wrote it equal to $0$, but what will negative factorial mean? $$^nC_{n+1} = \frac{n!}{(n+1)!\cdot(n-n-1)!} = \frac{n!}{(n+1)!\cdot(-1)!}$$ ...
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Birthday problem, I'm confused by this formula

I've given the following statement (n is given, and equals 100) : Now, I'm quite confused by the second binomial coefficient: how can that represent the days for the birthdays of the remaining 98 ...
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4answers
66 views

Evaluate $\int_{0}^{1}(1-x)^ndx$ by expanding the bracket.

I'd like to get a hint on this exercise. I believe I'm somewhat close to the answer. I used the binomial theorem to get: $\displaystyle\int_{0}^{1}(1-x)^ndx = \int_{0}^{1}\sum_{k=0}^{n}{n\choose ...
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1answer
29 views

Show that $r_k^n/n \le \binom{kn}{n} < r_k^n$ where $r_k = \dfrac{k^k}{(k-1)^{k-1}}$

Show that for $n \ge 2$, $\dfrac{r_k^n}{n+1} \le \binom{kn}{n} < r_k^n$ where $r_k = \frac{k^k}{(k-1)^{k-1}}$. This is a generalization of How to prove through induction which asks for a proof ...
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Application of the Binomial Theorem-remainder

I am having a confusion in this question- What is the remainder when $7^{103}$ is divided by 24? I attempted it as follows - It can be written as $(7^2)^{51} \cdot 7$ Which can be written as ...
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How to compute linear recurrence of a sum of binomial-multiplied linear recurrences [duplicate]

I have $$g(n) = \sum_{k=1}^{n} \binom{n}{k}f(k)$$ where $f(k)$ is a large linear recurrence. $g(n)$ is also a linear recurrence as well. Normally, when computing the value of a linear recurrence, I ...
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Simplying linear recurrence sum with binomials

Is there a way to simplify $$\sum_{k=1}^{n} \binom{n}{k}f(k)$$ Where $f(k)$ is a large linear recurrence?
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Coefficients for the falling factorial

Hello fellow mathematicians, I am trying to find a generating function, or at least find some useful property from the coefficients of the falling factorial. Let $(x)_n$ denote a falling factorial, ...
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Orthogonality for Binomial Coefficients

Could somebody explain to me where these two formulas come from as applications of the binomial theorem? $$\sum_{k=0}^n {n \choose k}(-1)^kk^r=0$$ for non-negative integers $r\lt n$. And ...
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Weighted sum of binomial coefficients by powers of lower value.

I am trying to calculate $\sum_{i=0}^n i^K {n \choose i}$ for $K \in \mathbb{N}$. Clearly, the case $K=0$ is trivially $2^n$ by the binomial theorem. For higher $K$ I am stumped. I know I can use: ...
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Binominial Theorem proving

As I was trying to understand the proof of Binomial Theorem by induction, I got stuck at this line. What formulas should be used to get from left to right part? Any explanations and answers ...
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1answer
20 views

Cofficient of x in a product

How can I efficiently find the coefficient of $x^m$ in the following product - $\prod\limits_{i=1}^{n-1}(1 - p_i + p_ix)$
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4answers
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Show that $\binom{2n}{n}$ is an even number, for positive integers $n$.

I would appreciate if somebody could help me with the following problem Show by a combinatorial proof that $$\dbinom{2n}{n}$$ is an even number, where $n$ is a positive integer. I ...
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1answer
50 views

Sum of Some Binomial Terms Equals Zero

Let $q$ and $\ell$ be positive integers. Then the sum $$ \sum_{k=q}^\ell (-1)^{k+q}\binom{k}{q}\binom{\ell}{k} = \left\{\begin{array}{ccc} 1 \mbox{ if } \ell =q\\ 0 \mbox{ if }\ell ...
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1answer
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Bound on the Beta function

For positive integers x and y, we have that $$ B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = \frac{1}{x} \left( \begin{array}{c} x+y-1 \\ x \end{array} \right)^{-1} . $$ However, $$ \left( ...
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2answers
61 views

Summation over a product of binomial coefficients

Question: I can't figure out why the following equality is true $\sum_\limits{k=a-b-c}^{d} (-1)^k \binom{d}{k}\binom{k+b+c}{a} = (-1)^d \binom{b+c}{a-d} $ How can this be shown? (In the book it just ...
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Stirling or better binomial approximation

Is there a method to find a approximation to $\log_2 \dbinom{n}{n^a}$ with $a\in(0,1)$? Similar to Approximating the logarithm of the binomial coefficient however here argument scales as radical of ...
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Show that $\binom{n}{k}=\frac{1}{2i\pi}\int_{C}\frac{(1+z)^{n}}{z^{k+1}}dz.$

I would like to prove that $$\binom{n}{k}=\frac{1}{2i\pi}\int_{C}\frac{(1+z)^{n}}{z^{k+1}}dz.$$ C is the circle at $0$ with radius $r>0$. I cannot get that expression, if I write the integral as ...
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2answers
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Why General Leibniz rule and Newton's Binomial are so similar?

The binomial expansion: $$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$ The General Leibniz rule (used as a generalization of the product rule for derivatives): $$(fg)^{(n)} = ...
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Combinatorial proof of $ \sum \limits_{i = 0} ^{m} 2^{n-i} {n \choose i}{m \choose i} = \sum\limits_{i=0}^m {n + m - i \choose m} {n \choose i} $

I've been wondering for a while how to solve (prove) a combinatorial identity, using just combinatorial interpretation: $$ \sum \limits_{i = 0} ^{m} 2^{n-i} {n \choose i}{m \choose i} = ...
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1answer
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How can find this two sequence recursive relations?

Let $$D_{n}=\sum_{j=0}^{n-1}(-1)^{n+j-1}\dfrac{\binom{2n-4}{j}}{n+j-1},E_{n}=\sum_{j=0}^{n-1}(-1)^{n+j-1}\dfrac{\binom{2n-4}{j}}{n+j}$$ I want find $D_{n}$ and $E_{n}$ recursive relations, I ...
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Is there limitation when applying binomial theorem?

Problem as title showed. $(a+b)^{-n}$. If $n$ is a positive integer. Can $a$ or $b$ be a complex number? Many thanks in advance.
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Evaluating $\sum_{0\leq k,l \leq n}\binom{n}{k}\binom{k}{l}l(k-l)(n-k)$ algebraically

I'm having problems with the following sum: $$\sum_{0\leq k,l \leq n}\binom{n}{k}\binom{k}{l}l(k-l)(n-k)$$ It's quite easy to think about it combinatorically: We have $n$ balls, we're coloring $k$ ...
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A sum expressed by a Kampe de Feriet function.

Let $a_1$,$a_2$, $a_3$ and $b_1$,$b_2$, $b_3$ be real numbers subject to $1+b_1+b_2 - b_3 > 0 $. By generalizing the result from A sum involving a ratio of two binomial factors. we have shown that ...
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Second question about a limit.

Is the following sequence converge? $$ \lim_{n\rightarrow\infty}\frac{1}{(1+M)^{2n}}\sum_{i=0}^{n}\left( \begin{array}{c} 2n ...
4
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2answers
49 views

What is wrong with ${13 \choose 1}{4 \choose 2} \cdot {12 \choose 1}{4 \choose 2}$ as combinations for two pair in poker?

Let's consider two pairs in a 52 cards deck of poker where every person gets five cards. My idea to approach this problem is to take following steps: First pair There are ${4 \choose 2}$ ...
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2answers
66 views

Binomal theorem show that $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\dots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\dots=2^{n-1}$

I'm having some trouble with this question Show that $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\dots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\dots=2^{n-1}$$ Attempt: Expanding $(1+1)^n=2^n$ ...