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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How can I show that $\sum_{k=0}^n [\frac{(n-2k)}{n}{n \choose k}]^2 = \frac{2}{n}{2n - 2 \choose n - 1}$?

I want to show this $\sum_{k=0}^n [\frac{(n-2k)}{n}{n \choose k}]^2 = \frac{2}{n}{2n - 2 \choose n - 1}$ I tried to solve it using this identities: I) $\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$ II) ${n \choose k} = {n-1 \choose…
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Find equation of Pascal variation to calcuate the nth element of x row

I built a Pascal Triangle, but instead of each value in a row being a sum of the two numbers above it I made each value the sum of the numbers above it divided by the product of those numbers. So, as an example the first three lines are 1, 1 1, 1 2…
mkinson
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Explain why twice the sum $\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}$ is $2^{12}-\binom{12}6$

Can someone explain how is the RHS concluded? I did with sample numbers and it is all correct. but I can't figure out how C(12,6) comes to play. $$ \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} =…
Sanone
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Alternating sum of binomial of coefficients up to $\binom{n}{r}$

Question : Evaluate - $$ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3} \dots + (-1)^r\binom{n}{r}$$ I tried to solve it but I am able to solve only when $r=n$.In that case I can put $x=-1$ in the expansion of $(1+x)^n$ and the result is $0$. I…
Jaideep Khare
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Coefficent of an equation

I am suppose to find the coefficent of x^2 in this equation after doing the calculation I ended up with this but that is the wrong answer what is that I am missing?
JackyBoi
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finding $\mathop{\sum\sum}_{0 \leq i < j\leq n}(i+j)\binom{n}{i}\binom{n}{j}$

finding $\displaystyle \mathop{\sum\sum}_{0 \leq i < j \leq n}(i+j)\binom{n}{i}\binom{n}{j}$ expanding sum $\displaystyle (0+1)\binom{n}{0}\binom{n}{1}+(0+2)\binom{n}{0}\binom{n}{2}+\cdots \cdots…
DXT
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Coefficient of $x^2$ in $(x+\frac 2x)^6$

I did $6C4 x^2\times (\dfrac 2x)^4$ and got that the coefficient of $x^2$ is $15$, but the answer is $60$, why? Did I miss a step?
JohnFire
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Binomial theorem expansion. Why do the powers flip in certain circumstances?

Okay given an expression like $(x+y)^5$ we have terms $x^ky^{n-k}$ So we have ${5 \choose 0} x^0y^5 + {5 \choose 1} x^1y^4+ {5 \choose 2} x^2y^3+ {5 \choose 3} x^3y^2+ {5 \choose 4} x^4y^1+ {5 \choose 5} x^5$ However when we have $(3x^2+y)^5$ we…
K. Gibson
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Binomial coefficient; substitution?

How do we get this equality? \begin{eqnarray*} \binom{n+l-1}{l}q^l=\binom{-n}{l}(-q)^l \end{eqnarray*} I haven't worked a lot with the binomial coefficient, so I don't really have a feel for these things... I really hope someone would help me with…
Sha Vuklia
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The proof of binomial identity $\binom{j+r-1}{j}=(-1)^j \binom{-r}{j}$?

I'm trying to understand the following proof for $\binom{j+r-1}{j}=(-1)^j \binom{-r}{j}$. $$ \begin{align} \binom{j+r-1}{j}&=\frac{(j+r-1)(j+r-2) \cdots r}{j!}\\ &=(-1)^j \frac{(-r-(j-1))(-r-(j-2)) \cdots (-r)}{j!} \\&=(-1)^j \frac{(-r)(-r-1) \cdots…
mavavilj
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find n, if the coefficient of $x^2$ in $(3 + 2x)^n$ is 20412.

Find $n$, if the coefficient of $x^2$ in $(3+2x)^n$ is 20412 I solved it using binomial theorem and got $n(n-1)3^{n-2} = 10206$ The term like $n(n-1)3^{n-2}$, I am not sure how to solve it.
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Inverse Binomial Coefficient Lookup/Check

Is there a quick way of checking - either by hand or computationally - whether or not a given integer is a binomial coefficient? If there isn't, is there a website where one can check this easily?
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binomial theorem: find coef. xy

Given: $$ \left(x-\dfrac{1}{2y}\right)^8\left(x+\dfrac{1}{2y}\right)^4 $$ Using binomial theorem, what is the coefficient of xy in the expansion? I've tried to do it but I couldn't. Could you please help me with it?
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Solving $\binom{39}{5+2x}=\binom{39}{2x-2}$

I have this equation: $$\binom{39}{5+2x}=\binom{39}{2x-2}$$ And I don't know how to solve it. I've tried by the definition of combination but I get stuck. I get stuck here:
tobi
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Binomial - Find values of a, b and n

I'm very confused. Would we have to expand the brackets? If the first three terms of the expansion $(a-2x)^n$ are $1-16x+bx^2 - ...$ then find the values of $a$, $b$ and $n$? Thank you so much!