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
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Solve $\binom{x+2}{5} = 126$ for $x$

How can we solve for $x$ in the equation $\binom{x+2}{5} = 126$? By expanding, we get $(x+2)(x+1)x(x-1)(x-2) = 126(5!)$, and by trial and error, we can see $x = 7$, but how can we solve for this without trial and error? Also, how do we know only one…
Trts
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Binomial identity

I'd like to get a hint to prove the following identity: $$\tag{1}\sum_{\nu}(-1)^{\nu}\displaystyle \binom{a}{\nu}\binom{n-\nu}{r}=\binom{n-a}{n-r} .$$ The original statement reads "By specialization derive from (12.9) the identity..." where (12.9)…
r_31415
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Expansion of $ (a_1 + a_2 + \cdots + a_k)^n $

Is there an expansion for the following summation? $$ (a_1 + a_2 + \cdots + a_k)^n $$
user2468
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Binomial expansion, how to do them quickly?

I'm currently preparing for a test where I'm bound to do a couple of binomial expansions. Since I never encountered them in my formal education, I looked how to do them myself and found out: $ (x+y)^n = \displaystyle \sum_{k=0}^n \binom{n}{k} x^k…
Phaptitude
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Simplify $\sum_{k=1}^{n} {k\choose m} {k}$

$\sum_{k=1}^{n} {k\choose m} {k}$ I have tried to expand it, but the m is pretty annoying. Any ideas to get rid of the summation and give a simple formula? There is a part before $\sum_{k=1}^{n} {k\choose m} {\frac{1}{k}}$, see if it gives any…
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Partial sum over $M$, of ${m+j \choose M} {1-M \choose m+i-M}$?

Is there (likely to be) any formula for $$ \sum_{m' \geq m} {m+j \choose m'} {B - m' \choose m+i-m'} $$ ? I am mainly interested in the case $B=1$ (or a prime power), if that's any easier. EDIT: as suggested, here are the $B=1$ answers for…
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Bounds for $\binom{n}{cn}$ with $0 < c < 1$.

Are there really good upper and lower bounds for $\binom{n}{cn}$ when $c$ is a constant $0 < c < 1$? I know that $\left(\frac{1}{c^{cn}}\right) \leq \binom{n}{cn} \leq \left(\frac{e}{c}\right)^{cn}$.
user35671
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The coefficient of $x^{101}$ in the expansion of $(1-x)(1-2x)(1-2^2x)...(1-2^{101}x) =?$

The coefficient of $x^{101}$ in the expansion of $(1-x)(1-2x)(1-2^2x)...(1-2^{101}x) =?$ Can I get some help please?Thanks in advance
humto
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a problem about binomial coefficients sum

I want to know how to caculate $\binom{2n}{0}^3-\binom{2n}{1}^3+\cdots+(-1)^k\binom{2n}{k}^3 \cdots+\binom{2n}{2n}^3$? The sum equals $ (-1)^{n}\binom{3n}{2n}\binom{2n}{n} $, but I donot know how to get this. thanks
wxu
  • 6,671
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Sum of Binomial coefficients identity

I am trying to find an exact formula for the following: $\sum\limits_{i=0}^{n}{\binom{2n}{n-i}\frac{2i^2+i}{n+i+1}}$ I don't think this should be too bad with a rearrangement of terms, but I keep getting stuck.
Joe
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Showing two definitions of a binomial coefficient are the same

I have a homework question where we have to prove the following definitions of a binomial coefficient are equal, algebraically. This is what I got so far, and it's getting pretty complicated. And I could use some directions on how to continue. At…
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lacunary sum of binomial coefficients

I am sure there must be a known estimate for sums of the form: $$ S_d(n)=\sum_{j=0}^n \binom{dn}{dj} $$ where $d\ge 1$. For $d=1$, the answer is obvious. For $d=2$, the answer is here: Sum with binomial coefficients: $\sum_{k=0}^{n}{2n\choose…
user28425
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Generalized Case: Three Consecutive Binomial Coefficients in AP

This is a generalization of my earlier question here posted recently, and is a more interesting one. Three consecutive binomial coefficients $$\binom n{r-1},\binom nr, \binom n{r+1}$$ are in an AP (arithmetic progression) with positive common…
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How to compute the sum $\sum_{r=0}^n \frac{(-1)^r}{{n \choose r}}$

Consider the sum $$\sum_{r=0}^n \frac{(-1)^r}{{n \choose r}}.$$ I know the sum is zero when $n$ is odd (pretty simple). The sum is $2-\frac{2}{2 + n}$ when $n$ is even. Can somebody provide a proof in the even case? Thanks
v kamat
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How can I simplify this expression involving binomial coefficients?

How can I simplify the following expression? $$\sum_{k=1}^n \binom{n}{k}^2$$
user5198
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