Mathematics Stack Exchange
Mathematics Stack Exchange

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: $$\binom{\alpha}{k}=\frac{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-k+1)}{k(k-1)(k-2)\dots1}$$

7695 questions
43
votes
2 answers

Proof for the upper bound and lower bound for binomial coefficients.

I have seen the bounds $\left(\frac{n}{k}\right)^k \leq {n \choose k} \leq \left( \frac{en}{k}\right)^k$ for integers $n \geq k >0$ for the binomial coefficient. I can prove the upper bound in this inequality, but I'm missing some detail for the…
user1105
  • 531
32
votes
4 answers

"Binomial theorem"-like identities

There are several identities which resemble the binomial theorem. For starters, we have the binomial theorem itself: $$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$$ But I just learned from the book "Concrete Mathematics", Exercise 5.37, that…
Ted
  • 33,788
25
votes
3 answers

I tried to kill the central binomial coefficients, but they came back.

Pascal's triangle contains huge numbers. I tried to "tame" it (make the numbers smaller) by "killing" the largest numbers, i.e. the central binomial coefficients: $2,6,20,70,252,924,\dots$. That is, as I built the rows (each number being the sum of…
Dan
  • 22,158
21
votes
2 answers

Negative Exponents in Binomial Theorem

I'm looking at extensions of the binomial formula to negative powers. I've figured out how to do $n \choose k$ when $n < 0 $ and $k \geq 0$: $${n \choose k} = (-1)^k {-n + k - 1 \choose k}$$ So now let's look at one case for using the binomial…
Tianxiang Xiong
  • 519
17
votes
3 answers

Binomial coefficients $1/2\choose k$

I don't understand questions that involve a binomial expression where you have a fraction choose $k$ or a negative number choose $k$. I understand and am able to do it when there are no fractions and they are all positive. We learned the generalized…
Tagwoh
  • 173
13
votes
1 answer

Sum of product of binomial coefficient

Is the following true? $$\sum_{x_1+x_2+...+x_n=n}\ \ \, \prod_{i=1}^{n}{k_i\choose x_i}={\sum_{i=1}^{n}k_i \choose n} .$$ I tried to use the multinomial theorem, but it doesn't seem applicable.
user14108
11
votes
2 answers

How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$

How prove this $$\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$$ This equation How prove it? Thank you I want take this $$f(x)=(1-x)^{n+1}?$$ But I can't deal this…
user94270
11
votes
2 answers

Complicated sum with binomial coefficients

I know how to prove, that $\frac{1}{2^{n}}\cdot\sum\limits_{k=0}^nC_n^k \cdot \sqrt{1+2^{2n}v^{2k}(1-v)^{2(n-k)}}$ tends to 2 if n tends to infinity for $v\in (0,\, 1),\ v\neq 1/2$. This can be proved with the use of Dynamical systems reasonings,…
Makar
  • 111
10
votes
3 answers

How prove this $\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $

I think the following equality is true ($p\in \mathbb{N},p\ge 2$): $$\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $$ when $p=2$, then $$\sum_{k=0}^{n}\binom{n}{k}^2\binom{2n+k}{k}=\binom{2n}{n}^2$$ But I can't prove…
math110
  • 93,304
9
votes
1 answer

How can I express $\sum_{k=0}^n\binom{-1/2}{k}(-1)^k\binom{-1/2}{n-k}$ without using summations or minus signs?

How can I express $$\sum_{k=0}^n\binom{-1/2}{k}(-1)^k\binom{-1/2}{n-k}$$ without using summations or minus signs?
Mec
  • 5,257
9
votes
4 answers

Evaluation of $\sum^{19}_{k=1}\frac{(-1)^{k-1}k}{\binom{20}{k}}$

Evaluating value of $$\sum^{19}_{k=1}\frac{(-1)^{k-1}k}{\binom{20}{k}}$$ I am trying to solve it using Integration. $$\int^1_0 t^{k}(1-t)^{n-k}dt=\frac{1}{n+1}\cdot \frac{1}{\binom{n}{k}}$$ So our sum is $\displaystyle…
jacky
  • 5,194
9
votes
3 answers

Value of $\displaystyle \frac{p}{q}$ in binomial expression.

If $\displaystyle p=\sum^{r}_{k=0}\binom{n}{2k}\binom{n-2k}{r-k}$ and $\displaystyle q=\sum^{n}_{k=r}\binom{n}{k}\binom{2k}{2r}\bigg(\frac{3}{4}\bigg)^{n-k}\bigg(\frac{1}{2}\bigg)^{2k-2r}$ , where $n\geq 2r$. Then value of $\displaystyle…
Priti Bisht
  • 421
9
votes
1 answer

Binomial sum gives $4^n$

I was looking at this question:Swapping the $i$th largest card between $2$ hands of cards and WolframAlpha gave me this result. Why is it so? $$\sum_{k=0}^n{2k\choose k}{2n-2k\choose n-k}=4^n?$$
Empy2
  • 50,853
8
votes
1 answer

Why does this sum of binomial coefficient ratios equal 1?

In the course of doing some calculations comparing unrepeatable sets of event trials, I ended up with the following identity. If my reasoning and my math are correct then this ought to be true, and it is indeed true for several values of $k$ and…
user150200
  • 111
8
votes
2 answers

Evaluation of ratio of two binomial expression

If $\displaystyle A = \sum_{k=0}^{24}\binom{100}{4k}.\binom{100}{4k+2}$ and $\displaystyle B = \sum_{k=1}^{25}\binom{200}{8k-6}.$ Then $\displaystyle \frac{A}{B}$ $\bf{My\; Try::}$ For evaluation of $$A=…
juantheron
  • 53,015
1
2 3
…
36 37