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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Find the coefficient of a term in the expansion of an algebraic expression

The coefficient of $x^3$ in the expansion of (1 + 2$x$ + 3$x^2$)$^6$ is equal to twice the coefficient of $x^4$ in the expansion of $(1 - a x^2)^5$. Find all possible value of constant $a$. I am actually getting $a=8$ or $a=-8$ using the binomial…
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Proof that binomial $\binom{n}{k}$ is an integer for $0>n\in\mathbb{Z}$

I have already looked at this proof here Proving that $n \choose k$ is an integer However I don't understand how I can use the Pascal identity for binommial coefficients if $n$ is a negative number. I have had the idea to prove a Connection between…
New2Math
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value of $f(2019)$ in binomial expression

if $\displaystyle f(n)=\mathop{\sum}_{i>j\geq 0}\binom{n+1}{i}\binom{n}{j}.$ Then value of $f(2019)$ is what i try $\displaystyle f(n)=\binom{n+1}{1}\binom{n}{0}+\binom{n+1}{2}\bigg(\binom{n}{0}+\binom{n}{1}\bigg)+\cdots +…
jacky
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ratio of binomial expression

sum of expression $$\large\frac{\sum^{r}_{k=0}\binom{n}{k}\binom{n-2k}{r-k}}{\sum^{n}_{k=r}\binom{2k}{2r}\bigg(\frac{3}{4}\bigg)^{n-k}\bigg(\frac{1}{2}\bigg)^{2k-2r}}(n\geq 2r)$$ $(a)\;1/2\;\;\;\;\;\; (b)\;2\;\;\;\;\;\; (c)\; 1\;\;\;\;\;\; (d)\;$…
jacky
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Sum of series having binomial coefficients

Prove that $\displaystyle \sum_{r=0}^n {n+r\choose r} \frac{1}{2^{r}}= 2^{n}$ what i try $$\binom{n}{n}+\binom{n+1}{1}\frac{1}{2}+\binom{n+2}{2}\frac{1}{2^2}+\binom{n+3}{3}\frac{1}{2^3}+\cdots…
jacky
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How to simplify this binomial coefficients?

For $a, b, r, n \in \mathbb N$ : I have to simplify $$\sum_{k = 0}^{r}\binom {a}k\binom {b}{r-k}$$ I tried by using factorials but it seemed more complicated
Lamethyste
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Is it possible to show

As part of a larger proof I need the following to be true in order to make it work, $(a+1){b\choose a+1}x^{a+1}(1-x)^{b-a-1}=a{b\choose a}x^a(1-x)^{b-a}$ Can anyone give me a hand?
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Serge Lang - Basic Mathematics - Sum of Binomial coefficients p.387 ex. 9

I can not understand the solution.How does the second step come about? How does the expression in brackets appear? solution image
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Solving for $n$

$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2} = 22$$ I'm trying to solve this equation for $n$. If $$\binom{n}{0} = 1$$ Then we have that $$1+\binom{n}{1}+\binom{n}{2} = 22 \implies \binom{n}{1}+\binom{n}{2} = 21$$ However, I'm stuck there. Could you…
Mark
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Show that $\sum_{k=o}^n \binom{n}{k}^2 = \binom{2n}{n}$

There is a hint for this ex.: using symmetry. I would appreciate another hint to take advantage of. My approach so far: Induction step: $\sum_{k=0}^{n+1} \binom{n+1}{k}^2$ $= 1 + \sum_{k=1}^{n+1} \binom{n+1}{k}^2$ $= 1 + \sum_{k=0}^{n}…
RM777
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Evaluating constant term

Consider you are given following $$\biggr (x-\dfrac{2}{x^2}\biggr )^6$$ I'm trying to evaluate the constant term. What I've done so far is given below $$\sum^{6}_{n = 0} \binom{6}{r}x^{6-r}\times (-2)^6 \times x^{-12}$$ $$\sum^{6}_{n = 0}…
Mark
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Finding the sum of Binomial Coefficients

The question says to find the value of $$\binom{n}{1} \cdot \left( \sum ^1 _ {r=0} \binom{1}{r}\right) + \binom{n}{2} \cdot \left( \sum ^2 _ {r=0} \binom{2}{r}\right) + \binom{n}{3} \cdot \left( \sum ^3 _ {r=0} \binom{3}{r}\right) \ldots…
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How to show $(\frac{n}{m})^m\leq\binom{n}{m}$?

How can I show that $$\left(\frac{n}{m}\right)^m\leq\binom{n}{m},$$where $m \leq n,$ and $m,n \in \Bbb N $?
ShaoyuPei
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For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that $x^n + (x+1)^n = (x+2)^n$.

For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that: $$x^n + (x+1)^n = (x+2)^n$$ Writing it using Newton's binom, we obtain: $$x^n = \sum_{i=1}^{n} \binom{n}{i} \cdot x^{n-i} \cdot (2^i - 1)$$ I don't know how to…
gareth618
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Calulating coefficiencies

I'm trying to calculate the number ofcoefficiencies of every $x^k$ in the expansion of $(x-\frac{1}{x})^{100}$ for an arbitrary $k\in\mathbb{Z}$ Now I tried the following formula: $\binom{100}{k}$ eventhough similar to the demonstration,…