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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Odd Multiplicities in Pascal's Triangle

This is more of a knowledge-sharing post, but I would love more insight. I was looking at Pascal's triangle and I was curious which numbers appear precisely an odd number of times. My first observations were: 1 appears infinitely often, so it is…
Zeda
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This question involves Pascal's Triangle & Binomial Theorem. Full Question

Could anyone please help me on the following problem: Factorize the expression $P(n)=n^x-n$ for $x=2,3,4,5$ Determine if the expression is always divisible by the corresponding $x$. If divisible use mathematical induction to prove your results by…
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Adding and subtracting values that don't exist

Above is a part of calculation involving the binomial coefficient. I understand the steps. What we have done is that we added and subtracted the same terms (those drawn in a red box) so that nothing changed by this step. But these terms don't exist…
user316849
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Formula for $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots$

I was doing an exercise about the integral $$ I_{n} = \int_{0}^{1} (1-x^{2})^{n}\, \mathrm{d}x $$ and I tried two approaches. On the one hand, by substituting $x = \sin u $ I found that $$ I_{n} = \frac{2}{3} × \frac{4}{5} × \cdots × \frac{2n}{2n+1}…
Shai
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Finding value of $\left \lfloor \frac{2017!}{2016!+2015!+\cdots \cdots + 2!+1!}\right\rfloor $

Finding value of the following $\displaystyle \bigg \lfloor \frac{2017!}{2016!+2015!+\cdots \cdots + 2!+1!}\bigg \rfloor $ Attempt : $$2017! = 2017\cdot 2016! = (1+1+1+\cdots \cdots +1)2016!>2016!+2015!+2014!+\cdots +2!+1!$$ Could some help me how…
DXT
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Sum of binomial coefficients involving $n,p,q,r$

Sum of binomial product $\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}$ Simplifying $\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}$. Could some help me…
DXT
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Parameters of a Binomial Coefficient

For a binomial coefficient $$\binom ab$$ would it be correct to say the following: $b$ must be either $0$ or a positive integer. i.e. $b$ cannot be negative or a fraction. $a$ can be either positive or negative, and either an integer or a…
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Binomial Coefficients with fractions: $\binom {m-\frac 12}m=\frac 1{2^{2m}}\binom {2m}m$.

From my earlier question here and the interesting solutions posted, we find interesting equivalents converting binomial coefficients with fractions to those without, e.g. $$\binom {m-\frac 12}m=\frac 1{2^{2m}}\binom {2m}m$$ and $$\binom {n+\frac…
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Using Binomial Theorem to prove identity

I need to prove the following using the binomial theorem $${n \choose k} = {n-2 \choose k} + 2{n-2 \choose k-1} + {n-2 \choose k-2}$$ The binomial theorem states $$(1+x)^n = \sum_{k=0}^n {n \choose k} x^k$$ I'm trying to use $(1+x)^n =…
Thatdude1
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${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as which of the following?

${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as ? A. ${21 \choose 5}$ B. ${20 \choose 5}-{11 \choose 4}$ C. ${21 \choose 5}-{10 \choose 5}$ D. ${20 \choose 4}$ Please give me a hint. I'm unable to group the…
Bazinga
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Solving a certain binomial sum involving the floor function

Does anybody know of any techniques to solve the following sum (or a fast way (polynomial in N) to compute it): $$\sum_{i=0}^{N} \binom{N-i}{A} \cdot \binom{\lfloor iP/Q \rfloor + 1}{B}$$ where $P$ and $Q$ are relatively prime positive integers, and…
Ryan Harris
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closed form expression for the sum of the first s items of alternating binomial coefficients

Is there a closed form expression for the following sum: $$\sum_{k=0}^s (-1)^k {n \choose k}, $$ where $s \in \{0,1,2,...,n\}$ -- which is basically the first $s$ terms in the alternating binomial coefficients series. I know if $s=n$, the sum is…
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Binomial coefficients based question

$\binom {m}{0}+\binom {m}{1}-\binom {m}{2}-\binom {m}{3}+\binom {m}{4}+\binom {m}{5}-\binom {m}{6}-\binom {m}{7}+........=0$ if and only if for some positive integer k,then $m=$ $(A)4k\hspace{1cm} (B)4k+1\hspace{1cm}(C)4k+2\hspace{1cm}(D)4k+3$ My…
Vinod Kumar Punia
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Another binomial coefficient sum

In my work I ran across the following binomial coefficient sum: $$ S=\sum_{a=0}^{n-1-l} (-1)^a \binom{n}{l+1+a} \binom{l+a}{l} $$ where $n\geq 0$ and $0\leq l \leq n-1$. I browsed the web and found (thanks to this question) the formula (5.24) in…
Stefan Hante
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max number of times integer $>1$ appears in pascals triangle

$120, 210 ,3003$ appear $6$ times in Pascal's triangle. $120={10\choose3}={16\choose2}={120\choose1}\\$ $210={10\choose4}={21\choose2}={210\choose1}\\$ $3003={14\choose6}={78\choose2}={3003\choose1}$ Are there any numbers $>1$ that appear more than…