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Friday, 07 March 2014 21:05

Binomial Coefficients

By  Gideon
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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:

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

We have already proved that:
\[ \sum_{i=1}^{n}{i} = \frac{(n+1)(n)}{2}. \]

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:
\[ \binom{10}{5} = \frac{10}{2} \frac{9}{3} \frac{8}{4} \frac{7}{1} \]
\[ = (5)(3)(2)(7) = 210. \]

Read 1553 times Last modified on Saturday, 08 March 2014 03:48
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