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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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Sum of binomial coefficients from $\binom{2m+1}{0}$ to $ \binom{2m+1}{m} $

Prove that: $\displaystyle 4^m = \binom{2m+1}{0}+\binom{2m+1}{1}+\binom{2m+1}{2}+\ldots + \binom{2m+1}{m} $ From the Binomial Theorem: $\displaystyle (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$ If $\displaystyle a = b = 2$, then we…
asd
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Sum of coefficients in binomial theory.

While trying to get introduced to binomial theory at university's website, I learned about the sum of binomial coefficients, and they showed me some of the features, and one of them was the pyramid of coefficients sum which is: $$1 = 1$$ $$1+1 =…
Firas Abd El Gani
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Proving a certain sum is 1

I'd like to prove that, for any $M,N\in\mathbb{N}$ with $M\leq N$, and any $n\in\mathbb{N}$ with $n\leq M$, the sum: $$\sum\limits_{k=0}^n\frac{\binom{M}{k}\!\!\binom{N-M}{n-k}}{\binom{N}{n}}=1.$$ I have proved it for $n=2$ by hand, and got stuck on…
MickG
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Find $k$ given that ${14 \choose k} = {14 \choose k-4}$

Ok, so I stumbled upon the question on the title these days, when going over Apostol's Calculus I. Now, because of the placement of the question in the exercises section, I'm convinced that the book wants the reader to use the well known identity…
ulilaka
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Find the sum of all coefficients in expansion of $(x^2+2x)^{20}$.

Find the sum of all coefficients in expansion of $(x^2+2x)^{20}$. Is there a specific method or formula for this?
Satvik Mashkaria
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How find this sum closed form?

Have this sum have close form $$f(n)=\sum_{k=0}^{n-1}\left(\left(\sum_{i=0}^{k}(-1)^i\binom{n}{i}\right)\cdot\left(‌​\sum_{j=k+1}^{n}(-1)^j\binom{n}{j}\right)\right)$$ Maybe this sum can use integral to solve it? Thank you
math110
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The asymptotic behavior of the CDF of Binomial distribution

I got stuck with the following problem which seemed not to be very complicated at the beginning! I would like to compute the CDF of a Binomial distribution, \begin{equation*} F(\ell;n,q) = \sum_{k=0}^{\ell} \binom{n}{k} q^k (1-q)^{n-k}…
MikeL
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How find this sum $\sum_{k=0}^{p}t^k\binom{n}{k}\binom{m}{p-k}$

Find the closed form $$\sum_{k=0}^{p}t^k\binom{n}{k}\binom{m}{p-k}$$ since $$\binom{n}{k}\binom{m}{p-k}=\dfrac{n!}{(n-k)!k!}\cdot\dfrac{m!}{(p-k)!(m-p+k)!}$$ then I can't
math110
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How find this sum $\sum_{k=0}^{\left[\frac{n}{2}\right]}\frac{(-1)^k\binom{n-k}{k}}{n-k}$

How find this sum $$\sum_{k=0}^{\left[\dfrac{n}{2}\right]}\dfrac{(-1)^k\binom{n-k}{k}}{n-k}$$ My try:since $$\dfrac{(-1)^k}{n-k}=\int_{-1}^{0}x^{n-k-1}dx$$ then I can't Thank you very much!
user94270
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Proof of identity involving binomial coefficients

I'll be happy if you could help me prove this argument with algebraic tools: $${N\choose 0}a^N+{N\choose 1}a^{N-2}+{N\choose 2}a^{N-4}+{N\choose 3}a^{N-6}+\dots = \frac{a^2+1}{a}$$ Thanks, Don
Don
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Concrete Mathematics, Newton Series and Inversion

In section 5.3 of Concrete Mathematics, on the bottom of page 192, "A special case of the rule (5.45) we've just derived for Newton's series can be rewritten in the following way:" (this is 5.48) $g(n) = \displaystyle\sum\limits_{k}^{}…
user135483
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Finding a simple expression - Binomial Theorem

How does one find a simple expression of the one below applying the binomial theorem: $$\sum_{k=1}^n k \cdot 2^k{ n \choose k}$$ Edit: $\frac{d}{dx}(x^k)=kx^{k-1}$ $(1 + x)^n = \sum_{k=0}^n { n \choose k}x^k = \sum_{k=0}^n { n \choose…
Mike
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Why does n choose k give the number of ways to choose k copies of a's in binomial theorem?

In the expression $(a+b)^n$, I understand that we are summing together all of the combinations of length $n$ involving $a$ and $b$. What I don't understand is why $n$ choose $k$ corresponds to the number of ways to choose k copies of $a$'s in each…
Princess Mia
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Find the coefficient of $x^m$ in the expansion $(1 + ax + bx^2)^n$

Find the coefficient of $x^m$ in the expansion of $(1 + ax + bx^2)^n$ One approach would be: Let $p = 1+ax$ and $q = bx^2$ Now expand $(p+q)^n$ and then expand $p$ and $q$ individually. But that is so clumsy. Can we derive a direct formula?
Abhinav Chauhan
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$\binom{n}{j} \cdot (n - j)$ when $j = n$

I am a bit confused about simplifying $\binom{n}{j} \cdot (n - j)$ when $j = n$. If we simply plug $j = n$, we get $\binom{n}{n} \cdot 0 = 0$. But if we first perform a cancellation of the $(n-j)$, w'd get $$ \frac{n!}{j!(n-j)!} (n-j) = \frac{n!}{j!…
David
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