# MandysNotes

Friday, 07 March 2014 21:05

## Binomial Coefficients

By  Gideon
Rate this item

Definition:

For any natural number $$n$$, and any natural number $$i$$, such that, $$n \geq i \geq 0$$:

The binomial coefficient,

$\binom{n}{i},$

of $$n$$ and $$i$$ is,

$\binom {n}{i} := \frac{n!}{(n-i)!(i)!}.$

From this definition, and from the fact that 0! = 1, we see that:

$\binom{n}{0} = \frac{n!}{(n-0)!(0)!} = \frac{n!}{n!(1)} = 1,$

and that:

$\binom{n}{1} = \frac{n!}{(n-1)!1!} = \frac{(n)(n-1)(n-2)...(1)}{(n-1)(n-2)...(1)} = n.$

The case $$\dbinom{n}{2}$$ for $$n \geq 2$$ is more interesting.
By using $$i = 2$$ in we see that:

$\binom{n}{2} = \frac{n!}{(n-2)!(2)!} = \frac{(n)(n-1)(n-2)...1}{(n-2)...(1)(2)!} = \frac{(n)(n-1)}{2},$

And in general:

$\binom{n}{i} = \frac{(n)(n-1)...(n-i+1)}{(i)!} .$

Replacing $$i$$ by $$(n-i)$$ in the definition of the binomial coefficient

(for $$n \geq i \geq 0$$), we obtain:

$\binom{n}{n-i} = \frac{n!}{((n - (n-i))!(n-i)!} = \frac{n!}{(i)!(n-i)!} = \binom{n}{i}.$

Therefore we have proved:

for every natural number $$n$$, and every natural number $$i$$ such that $$n \geq i \geq 0$$,
$\dbinom{n}{i} = \dbinom{n}{n-i}.$

Combining this with our previous results we see that:

$\binom{n}{n - 0} = \binom{n}{n} = \binom{n}{0} = 1,$

that:

$\binom{n}{n-1} = \binom{n}{1 } = n,$

and that:

$\binom{n}{n - 2} = \binom{n}{2} = \frac{(n)(n-1)}{2}.$

$\sum_{i=1}^{n}{i} = \frac{(n+1)(n)}{2}.$

Combining this with the result:
$\binom{n}{2} = \frac{n(n-1)}{2},$

implies:

$\sum_{i=1}^{n}{i} = \binom{n+1}{2}.$

Example Problem:

Find the sum of all natural numbrs from one to one hundred:

$\sum_{i=1}^{100}{i} = ?$

Solution:

$\sum_{i=1}^{100}{i} = \binom{101}{2} = \frac{(101)(100)}{2} = (101)(50) = 5050.$

Example Problem:

Find the numerical value of the binomial coefficient: $$\dbinom{10}{4}$$.

Solution:

$\binom{10}{4} = \frac{10!}{(10 - 4)!4!} = \frac{10!}{6!4!}$
$=\frac{(10)(9)(8)(7)}{(4)(3)(2)(1)} = \frac{10}{4}\frac{9}{3} \frac{8}{2} \frac{7}{1} .$
Rearranging terms give:
$\binom{10}{5} = \frac{10}{2} \frac{9}{3} \frac{8}{4} \frac{7}{1}$
$= (5)(3)(2)(7) = 210.$