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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Show $241|\binom{240}{120}$.

Here is what I found: $\binom{241}{120}*\frac{121}{241}=\binom{240}{120}$. I'm not sure how to continue.
user614287
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Given that the coefficients of the $6^{th}$ term and $16^{th}$ term in the Expansion of $(x + y) ^n$ are equal. Find the value of $n$.

The Expansion gives me: $6^{th}$ term = $\displaystyle\binom{n}{5} x^{n-5}y^5$ and $16^{th}$ term = $\displaystyle\binom{n}{15} x^{n-15}y^{15}$. What do I do next?
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Why does $\binom{2n}{n} = \prod_{i=1}^n (\frac{n+i}{i} )$

I cannot understand why are you able to generalize this like this $$\binom{2n}{n} = \frac{2n × (2n−1) × ... × (n+ 2) × (n+ 1)}{n ×(n−1)×...×2×1} =\prod_{i=1}^n (\frac{n+i}{i}) $$ I get that $$\binom{2n}{n} = \frac{2n!}{n! (n-n)!} $$ and therefore…
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If the sum of coefficients in the expansion of (1+2x)^n is 6561 then the greatest binomial coefficient in the expansion is?

I have managed to find out the general term as nC0+2*nC1+2^2nC2+....+2^n*nCn. How do I approach after this? Please help.
Akash Roy
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binomial coefficient aproximation

I found this aproximation $$ \binom{n}{k}\sim \frac{n^k}{k!} $$ But there was a note that said this holds just when $n>>k$, why is this? I need to consider this binomial coefficient when $k$ is near of $n/2$ and I would like use this or any similar…
Luis GC
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subtracting these two coefficients

I am given three natural numbers $n>x\geq k$ and trying to find the value of $$ \binom{n}{k}-\binom{n-x}{k} $$ I reached the following and not sure if I'm in the right…
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Simplify $C(n,k)/C(n+1,k-1)$

I need to simplify $C(n,k)/C(n+1,k-1)$ without the answer containing any factorials or binomial coefficients. I know the answer is $((n-k+1) (n-k+2))/((n+1) k)$. I just have no clue how to go about getting there.
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dividing $\binom{n-m-1}{k}$ by $\binom{n-1}{k}$

for $n,m\in \mathbb{N}$ and $m
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Prove that $\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^m=...$

Let $\binom{n}{k}$ denotes the number of subsets with $k$ elements in $n$-elements set. Prove that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^m=\begin{cases} 0, & \text{ if } 0\le m \le n-1; \\ 2^n \cdot n!, & \text{ if } m=n. \end{cases}$$
Roman83
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Closed form for a binomial identity

$\textrm {How do I find a closed form for } \sum_{j=0}^n{} j\displaystyle\binom{j}{r} = ?$ Is this some kind of upper index summation? Any previous papers? Thank you
frank
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Closed-form expression for $\binom{n}{1}+3\binom{n}{3}+5\binom{n}{5}+\cdots$

Find a closed-form expression for $$\binom{n}{1}+3\binom{n}{3}+5\binom{n}{5}+\cdots ,$$ where $n > 1$. You may find the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ helpful. I really can't use the identity, as there is a floor function…
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Equality involving binomial coefficient

I think the following equality is correct, but I'm not sure, so I'm asking you: $$\left(\prod_{\large\tfrac{n}{2}\,<\,p\,\le\,\tfrac{6n}{7}}p\right)\cdot\left(\prod_{\large n\,<\,p\,\le\,3n\strut}p\right)<\binom{3n}{\frac{n}{2}}$$ for even $ n $. $…
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Is there a short expression for this?

Is there a closed form expression for $$\Bigg(\binom{n}{k-1}+\binom{n-1}{k-1}+\dots+\binom{k-1}{k-1}\Bigg)(k-1)!= \sum_{i=0}^{n-k+1}\frac{(n-i)!}{(n-k+1-i)!},$$ $$\Bigg(\binom{n}{k-1}+\binom{n-1}{k-1}+\dots+\binom{k}{k-1}\Bigg)(k-1)!=…
Turbo
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Show that $(m-1)^{n-1}*n^{m-2}=\binom{n+m-1}{n}$

I need help proving that $(m-1)^{n-1}*n^{m-2}=\binom{n+m-1}{n}$.
Fryguy
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Find Pascal's Triangle Nth row Xth element

row 1 | 1 row 2 | 1 1 row 3 | 1 2 1 row 4 | 1 3 3 1 row 5 | 1 4 6 4 1 I recently made a computer program that involved Pascal's Triangles. However I got obsessed with optimizing it. I know there is a way to get Nth row / Xth element of…
Nick
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