Questions tagged [binomial-coefficients]

For questions involving the coefficients involved in the binomial theorem. $ \binom{n}{k}$ counts the subsets of size $k$ of a set of size $n$.

The binomial coefficient $\binom{n}{k}$ can be defined in several equivalent ways for $n$ and $k$ non-negative integers:

  1. The number of subsets of size $k$ of a set of size $n$.
  2. Element $k$ of row $n$ in Pascal's triangle (counting the first element or row as $0$).
  3. $\dfrac{n!}{k!(n-k)!}$
  4. The coefficient of $x^k$ in $(1+x)^n$.

The binomial theorem says that $$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k$$ using the convention that $0^0=1$.

Binomial coefficients can be extended for arbitrary complex $\alpha$ through the formula: $$\binom{\alpha}{k}=\frac{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-k+1)}{k(k-1)(k-2)\dots1}$$

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Can I get a light hint for this binomial proof?

Consider the geometric series $$S=1+(1+x)+(1+x)^2+\dots+(1+x)^n$$ (a) Show that $$S=\frac{(1+x)^{n+1}-1}{x}$$ (b) Hence show that $$S={n+1\choose1}+{n+1\choose2}x+\dots+{n+1\choose r+1}x^r+\dots+{n+1\choose n+1}x^n$$ (c) Hence prove…
user71207
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Inconsistency in binomial expansion?

When I do a binomial expansion on $\frac{1}{(1-2x)(1+3x)}$ about $0$, I can do it in 2 ways. Method 1 $$\frac{1}{(1-2x)(1+3x)}=\frac{1}{1-2x}\frac{1}{1+3x}$$ Thus, getting $(1+y_1)^{-1}(1+y_2)^{-1}$. This method would give me a radius of convergence…
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How do i reduce this expression of binomial coefficients

I was solving a problem and am stuck with this expression. Any leads on how can I simplify this expression? $$\frac{{\sum\limits_{x=Q}^{N-P+Q} (x-Q) \binom{x}{Q} \binom{N-x}{P-Q}}}{{\sum\limits_{x=Q}^{N-P+Q} \binom{x}{Q} \binom{N-x}{P-Q}}}$$ UPDATE:…
nims
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Continuing "Pascal's triangle" for negative binomial exponents

Does there exist a pattern for the coefficients in a negative binomial expansion? This question is not about the explicit formula, but rather if there exist a continuation for Pascal's triangle. $$\begin{array}l (a+b)^{-2} &=&&&&…
Frank Vel
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A sum involving binomial coefficients and powers of 2

I am interested in a simplified version of the following sum $$\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^k}{2^k-1}.$$ I have to evaluate it for values of n ranging from $10^{4}$ till $10^{10}.$ Is there a way to express it in terms of some special…
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Prove that Pascals triangle contains only natural numbers, using induction.

I'm currently working my way through Spivak, and I'm stuck on the following. Prove that Pascals triangle only contains natural numbers using induction and the following relation: $\left( {\begin{array}{*{20}c} n+1 \\ k \\ \end{array}} \right)=\left(…
user17137
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Negative binomial coefficient

For $r \geq 1$, $k \geq 0$ both integers, I wish to show that $$\binom{-r}{k}^{*}(-1)^{k} = \binom{r+k-1}{k}$$ (the negative binomial coefficient is the left one). By definition, $$\binom{-r}{k}^{*}(-1)^{k} =…
Clarinetist
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Intuitive explanation of binomial coefficient formula

Regarding the formula for binomial coefficients: $$\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$ the professor described the formula as first choosing the $k$ objects from a group of $n$, where order matters, and then dividing by $k!$ to adjust…
Mark
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Nested... binomials coefficients?

Can I have a proof that this number exists? The number: $$\binom{1}{\binom{2}{\binom{3}{\binom{4}{\vdots}}}}$$ If the number exists, then what is the closed form of that number?
user253055
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Binomial Coefficients in the Binomial Theorem - Why Does It Work Question

to keep it simple: Given $(a+b)^3=\binom{3}{0}a^3+\binom{3}{1}a^2b+\binom{3}{2}ab^2+\binom{3}{3}b^3$ Could you please give me an intuitive combinatoric reason to why the binomial coefficients are here? for instance, what does $\binom{3}{2}ab^2$ mean…
Anonymous
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Solve for $n$, where $n$ is a positive integer

I have $$ {n \choose 2} = 21 $$ and as the title mentions I have to solve for $n$, but so far all I have managed to get to is $$n^2 -n =42 $$ and from there I'm completely lost. Any hints would be greatly appreciated.
Abby
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Is there a simpler way to compute this sum?

For any given positive integers $m$, $n$ and $q$, such that $m\leq n$ the following sum $$S_p= \sum_{p=0}^m(-1)^{p+q} \binom mp \binom mq \binom np \binom nq\frac{p! q! (m+n-p-q)!}{m! n!}$$ is equal to $1$ if $q=0$ and $0$ otherwise. The result can…
Tom-Tom
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Evaluate $\sum\limits_{k=2}^n \frac{n!}{(n-k)!(k-2)!} $

Question is to Evaluate $$\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!} $$ What i have done so far is $$\sum_{k=2}^n \frac{n!}{(n-k)!(k-2)!}=n(n-1)\sum_{k=2}^n \frac{(n-2)!}{(n-k)!(k-2)!}=n(n-1)\sum_{k=2}^n \binom{n-2}{k-2}$$ I could see that $$\sum_{k=2}^n…
user87543
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Expansion concerning the binomial theorem

The question goes: Expand $(1-2x)^{1/2}-(1-3x)^{2/3}$ as far as the 4th term. Ans: $x + x^2/2 + 5x^3/6 + 41x^4/24$ How should I do it?
Sophia
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Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $

for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $ where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$ My try:: I have expand the expression $\displaystyle…
juantheron
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