## Converting the Formula for Combinations...

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.
maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
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### Converting the Formula for Combinations...

Hi,
Wasn't sure if this belonged in Intermediate or Beginning Algebra. Apologies if this is misplaced.

I've been playing around with the formula for Combinations (n!/(n-r)!*r!) and was trying to answer the question: from the set n of 26 letters of the alphabet, for what value of r is the possible number of sets the greatest.

At first I thought it would be either 1 or 26 but discovered that not to be the case. As I tried more values I discovered a symmetrical pattern around the midpoint of 13, which I thought was interesting. I then wanted to try to graph that on my graphing calculator but I'm struggling to convert the Combination formula above into an equation of the form y=ax^2 + bx + c, (where n=x, r=y) which is the form I'm assuming would be needed.

If that's the correct form needed to graph it then I'm struggling with doing the algebraic manipulation to get it into that form.

Am I missing something?

Martingale
Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Converting the Formula for Combinations...

maroonblazer wrote:Hi,
Wasn't sure if this belonged in Intermediate or Beginning Algebra. Apologies if this is misplaced.

I've been playing around with the formula for Combinations (n!/(n-r)!*r!) and was trying to answer the question: from the set n of 26 letters of the alphabet, for what value of r is the possible number of sets the greatest.

At first I thought it would be either 1 or 26 but discovered that not to be the case. As I tried more values I discovered a symmetrical pattern around the midpoint of 13, which I thought was interesting. I then wanted to try to graph that on my graphing calculator but I'm struggling to convert the Combination formula above into an equation of the form y=ax^2 + bx + c, (where n=x, r=y) which is the form I'm assuming would be needed.

If that's the correct form needed to graph it then I'm struggling with doing the algebraic manipulation to get it into that form.

Am I missing something?

${{n}\choose{r}}=\frac{n!}{r!(n-r)!}$

Then

${{n}\choose{r+1}}=\frac{n!}{(r+1)!(n-(r+1))!}=\frac{n!}{(r+1)!(n-r-1))!}=\frac{n!}{(r+1)r!(n-r-1))!}=\frac{n!(n-r)}{(r+1)r!(n-r-1))!(n-r)}=\frac{n-r}{r+1}\frac{n!}{r!(n-r))!}=\frac{n-r}{r+1}{{n}\choose{r}}$

so ${{n}\choose{r+1}}\geq{{n}\choose{r}}$ if $\frac{n-r}{r+1}\geq1$

maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
Contact:

### Re: Converting the Formula for Combinations...

Martingale wrote:
maroonblazer wrote:Hi,
Wasn't sure if this belonged in Intermediate or Beginning Algebra. Apologies if this is misplaced.

I've been playing around with the formula for Combinations (n!/(n-r)!*r!) and was trying to answer the question: from the set n of 26 letters of the alphabet, for what value of r is the possible number of sets the greatest.

At first I thought it would be either 1 or 26 but discovered that not to be the case. As I tried more values I discovered a symmetrical pattern around the midpoint of 13, which I thought was interesting. I then wanted to try to graph that on my graphing calculator but I'm struggling to convert the Combination formula above into an equation of the form y=ax^2 + bx + c, (where n=x, r=y) which is the form I'm assuming would be needed.

If that's the correct form needed to graph it then I'm struggling with doing the algebraic manipulation to get it into that form.

Am I missing something?

${{n}\choose{r}}=\frac{n!}{r!(n-r)!}$

Then

${{n}\choose{r+1}}=\frac{n!}{(r+1)!(n-(r+1))!}=\frac{n!}{(r+1)!(n-r-1))!}=\frac{n!}{(r+1)r!(n-r-1))!}=\frac{n!(n-r)}{(r+1)r!(n-r-1))!(n-r)}=\frac{n-r}{r+1}\frac{n!}{r!(n-r))!}=\frac{n-r}{r+1}{{n}\choose{r}}$

so ${{n}\choose{r+1}}\geq{{n}\choose{r}}$ if $\frac{n-r}{r+1}\geq1$

Wow. Ok, deep breath...

Question #1: Why do we need to do: ${{n}\choose{r+1}}$?

Question #2: I'm following you up until: $\frac{n!}{(r+1)!(n-r-1))!}$ but then I can't make out how you get from there to: $\frac{n!}{(r+1)r!(n-r-1))!}$

Specifically, how are you able to insert 'r' into the denominator before the first factorial without changing any of the other terms?

Apologies if this should be obvious.

Martingale
Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Converting the Formula for Combinations...

maroonblazer wrote:
Wow. Ok, deep breath...

Question #1: Why do we need to do: ${{n}\choose{r+1}}$?

Question #2: I'm following you up until: $\frac{n!}{(r+1)!(n-r-1))!}$ but then I can't make out how you get from there to: $\frac{n!}{(r+1)r!(n-r-1))!}$

Specifically, how are you able to insert 'r' into the denominator before the first factorial without changing any of the other terms?

Apologies if this should be obvious.

1)the maximum value of ${{n}\choose{r}}$ is when r is in the middle of n and zero

to show this...

we can show that ${{n}\choose{r}}$ is an increasing function of $r$ $\left(\text{so long as }\frac{n-r}{r+1}\geq 1\right)$

after that point ${{n}\choose{r}}$ is a decreasing function of $r$

2) $(r+1)!=(r+1)\cdot r!$

the same way that $6!=6\cdot5!$