Combinations Problem

[cooldog182]

Sorry about how I have formatted this but I have to show for all integers n, k>=1 that

(n+1 over k) = (n+1 / k)(n over k-1)

now, since

(n over r) = n! / r!(n-r)!

and

(n over r) + (n over r-1) = (n! / r!(n-r)!) + (n! / (r-1)!(n-r+1)!)

this also equals

= (n+1 over r)

I have been able to have a bit of a go at it.

--------------MY TRY-----------

(n+1)! / k!(n-k)! = (n+1/k) (n! / (k-1)!(n-k+1)!)

= (n+1)! / (k(k-1)!(n-k+1)!

and since

k(k-1)!= k!

then we really want k(n-k+1)! to equal (n-k)!

but as far as I can see, this isn't right.

What am I doing wrong?

[/tex]

 

[pka]

Cool it is hard to follow what you mean.
Try TeX.
\(\displaystyle {n \choose k} \) to get \(\displaystyle {n \choose k}\)

Is this the problem \(\displaystyle \left( \begin{array}{c}
n + 1 \\
k \\
\end{array} \right) = \frac{{\frac{{n + 1}}{k}}}{{\left( \begin{array}{c}
n \\
k - 1 \\
\end{array} \right)}}\)?

 

[cooldog182]

Cool it is hard to follow what you mean.
Try TeX.
\(\displaystyle {n \choose k}\) to get \(\displaystyle {n \choose k}\)

Is this the problem \(\displaystyle \left( \begin{array}{c}
n + 1 \\
k \\
\end{array} \right) = \frac{{\frac{{n + 1}}{k}}}{{\left( \begin{array}{c}
n \\
k - 1 \\
\end{array} \right)}}\)?

no the n+1/k is to the left of the (n over k-1)

I will practice latex and get back to you

 

[pka]

\(\displaystyle \L
\begin{array}{c}
\left( \begin{array}{c}
n + 1 \\
k \\
\end{array} \right) &=& \frac{{\left( {n + 1} \right)!}}{{k!\left( {n + 1 - k} \right)!}} = \frac{{(n + 1)}}{k}\frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}} \\
\&
\\
\left( \begin{array}{c}
n \\
k - 1 \\
\end{array} \right) &=& \frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}} \\
\end{array}\)

 

[cooldog182]

\(\displaystyle \L
\begin{array}{c}
\left( \begin{array}{c}
n + 1 \\
k \\
\end{array} \right) &=& \frac{{\left( {n + 1} \right)!}}{{k!\left( {n + 1 - k} \right)!}} = \frac{{(n + 1)}}{k}\frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}} \\
\&
\\
\left( \begin{array}{c}
n \\
k - 1 \\
\end{array} \right) &=& \frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}} \\
\end{array}\)

yeah this is right!

 

[cooldog182]

\(\displaystyle \L

\left( \begin{array}{c}
n+1 \\
k \\
\end{array} \right)



=


\frac{{n+1}}{{\left (k)\right}} \\\)

and then next to n+1/k there is another (n over k-1) but i'm having trouble with tex

 

[cooldog182]

\(\displaystyle \L

\left( \begin{array}{c}
n+1 \\
k \\
\end{array} \right)



=


\frac{{n+1}}{{\left (k)\right}} \\\)

and then next to n+1/k there is another (n over k-1) but i'm having trouble with tex

and this is what i'm trying to show by the way (for all integers n, k>=1)

 

[cooldog182]

\(\displaystyle \L

\left( \begin{array}{c}
n+1 \\
k \\
\end{array} \right)



=


\frac{{n+1}}{{\left (k)\right}} \\\)

and then next to n+1/k there is another (n over k-1) but i'm having trouble with tex

and this is what i'm trying to show by the way (for all integers n, k>=1)

WAIT! did I just answer my own question? This confuses me

Thanks for your insight!

 

[soroban]

Hello, cooldog182!

Show for all integers \(\displaystyle n,\;k\,\geq\,1\) that: \(\displaystyle \,\begin{pmatrix}n+1 \\ k\end{pmatrix} \:=\:\frac{n+1}{k}\cdot\begin{pmatrix}n \\ k-1\end{pmatrix}\)

The right side is: \(\displaystyle \L\,\frac{n+1}{k}\,\cdot\,\frac{n!}{(k-1)!(n-k+1)!} \;= \;\frac{(n+1)n!}{k(k-1)!(n-k+1)!}\)

. . . . . . . . \(\displaystyle \L\:= \;\frac{(n+1)!}{k!(n-k+1)!} \;= \;\begin{pmatrix}n+1 \\ k\end{pmatrix}\)