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
-2
votes
2 answers

If $n\ge 2$, find $\binom 31 -\binom 42 +\binom 53....(-1)^{n-1} \binom {n+2}{n}$

I have no clue on how to start solving. Can I get a hint to get me started?
Aditya
  • 6,191
-2
votes
2 answers

Alternating Binomial Series Summation.

Finding $\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}\cdots \cdots +(-1)^r\binom{n}{r}$ Try: using Identity $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$ So $$\sum^{r}_{k=1}(-1)^r\bigg(\binom{n-1}{k}+\binom{n-1}{k-1}\bigg)$$ Ciuld some…
DXT
  • 11,241
-2
votes
2 answers

Question on binomial coefficient

What will be the coefficient of $x^{99}$ in the following: $(x+1)(x+2)(x+3).....(x+100)$ Please help. I am stuck on this.
-2
votes
2 answers

is this correct? $\binom{n}{k}=\binom{n}{k-1}-\binom{n-1}{k-1}$?

My instructor wrote $$\binom{n}{k}=\binom{n}{k-1}-\binom{n-1}{k-1}$$ I doubt it is incorrect. Is it?
DSL
  • 1,359
-2
votes
3 answers

Coefficient of $x^{18}$ in $(1+x+x^2+......x^9)^3$ .

We have to find coefficientof $x^{18}$ in $(1+x+x^2+......x^9)^3$ . I wrote it as $\frac{x^9-1}{x-1}$ after that got stuck . In the source I could not understand anything .
-2
votes
1 answer

sum with coefficient similar to binomial

$\sum_{l=0}^{k/2} \frac{k!}{(k-2l)!l!}$ I want to calculate the above sum. Its form is similar to that of binomial but not same. How to calculate it? Am I in wrong path?(which means its solution is not relevant to binomial?)
kyle
  • 1
  • 1
-2
votes
3 answers

Prove algebraically that ${n\choose k}=\frac{n(n-1)...(n-k+1)}{1\cdot 2\cdot ...\cdot k}$

From the definition of binomial coefficient, $${n\choose k}=\frac{n!}{k!(n-k)!}\Rightarrow \frac{n!}{k!(n-k)!}=\frac{n(n-1)...(n-k+1)}{k!}$$ $$\Rightarrow \frac{n!}{(n-k)!}=n(n-1)...(n-k+1)$$ Could someone explain how to derive the last equation…
user300048
  • 1,147
-2
votes
2 answers

Sum of binomial coefficients multiplied by $k^2$

Show the method used to evaluate $\displaystyle\sum\limits_{k=1}^{12} {12\choose{k}}k^2$ The answer is $159744.$
Siddhartha
  • 1,770
  • 18
  • 38
-3
votes
1 answer

Coefficients of expression $(5x+12x^2+15x^3+8x^4)^{15}$

Find the coefficient of $x^5$ in the expression $(5x+12x^2+15x^3+8x^4)^{15}$
-3
votes
1 answer

Binomial coefficients identity

Prove this identity: $$\binom {N-1/2} {N-1}=\frac{N}{4^{N-1}}\binom{2N-1}{N-1}$$ I was not able to find the proof of this
1 2 3
36
37