# Tagged Questions

Coefficients involved in the Binomial Theorem. $\binom{n}{k}$ counts the subsets of size $k$ of a set of size $n$.

49 views

### Last digit of a number

I was currently solving a question of permutations and in that I had to find the total ways of something. The answer was ${8\choose 4}$ which has last digit $0$ . A random thought that came to my ...
36 views

### Binomial Expansion - Finding the term independent of n.

The coefficient of $x^2$ in the expansion of $\left(1 + \frac x5\right)^n$, where $n$ is a positive integer, is $\frac 35$ . $(i)$ Find the value of $n$. $(ii)$ Using this value of $n$, find ...
19 views

### Why is $\sum\limits_{k:|2k/l-m/l|\geq\epsilon/2}\frac{\binom{m}{k}\binom{2l-m}{l-k}}{\binom{2l}{l}}\geq 2e^{-\epsilon^2l/8}$

This is stated in an article on the uniform convergence of probabilities of events to their relative frequencies. The idea behind the question is that I have a measurement on a sample of size $2l$ ...
34 views

55 views

### Binomial coefficient definition

Why is the definition of the binomial coefficient $${{m}\choose {r}}=\frac{m(m-1)\cdots(m-r+1)}{r!}$$ I'm not sure where the last term in the numerator came about. Why should there be a $+1$? ...
51 views

### Problem with binomial coefficients?

I am trying to find the sum of $$\sum_{x=0}^{n-2}\left (\frac{1}{x+1}{2x \choose x} \cdot \frac{1}{n-x-1}{2n-2x-4 \choose n-x-2}\right)\;.$$ I am told the answer is $$\frac{1}{n}{2n-2 \choose n-1}$$ ...
47 views

57 views

### Is Binomial:Gamma ever an integer?

We consider: $$\dfrac{\Gamma(n)}{\Gamma(k)\Gamma(n-k)}\quad\quad[1]$$ for $n,k\in\mathbb{R}$. Is $[1]$ ever an integer, except for the obvious?
56 views

42 views

### Evaluate $\binom{m}{i} - \binom{m}{1}\binom {m-1}{ i} + \binom{m}{2}\binom{m - 2}{i} - \ldots + (-1)^{m-i} \binom{m}{m-i}\binom{ i }{i}$

Evaluate the expression $$\binom{m}{i} - \binom{m}{1}\binom {m-1}{ i} + \binom{m}{2}\binom{m - 2}{i} - \ldots + (-1)^{m-i} \binom{m}{m-i}\binom{ i }{i}$$ I'm really stumped about trying to get ...
### How can I calculate $\sum_{k=1}^{n-1}\binom{n-1}{n-k}$?
I would like to know if I can calculate a closed expression for $$\sum_{k=1}^{n-1}\binom{n-1}{n-k}$$ This sum is equals to: $$1+(n-1)+(n-1)(n-2)+(n-1)(n-2)(n-3)+\ldots+(n-1)(n-2)/2+(n-1)$$