Discrete Math Proof.

[rainysomber]

I am slightly confused as to how to go about this.....

For a circle with n points on the perimeter, please do the following.

a) Determine a formula to count the number of quadrilaterals that can be formed by connecting these points.
b) Prove that your formula is correct.

I know C(n, 4) will work to solve this, but we need to create our own and pascal's formula is out. I know the recursive formula for cominationa s well, but not exactly how to obtain it.


Thanks for the help! = ]

 

[rainysomber]

I should say that the recursive formula I have found is

[(n-1) choose 3] multiplied by (n/4).

 

[soroban]

Hello, rainysomber!

For a circle with \(\displaystyle n\) points on the perimeter,

a) Determine a formula to count the number of quadrilaterals that can be formed by connecting these points.

If you're seeking a recursive formula . . .

\(\displaystyle \text{We want to go from: }\;a_n \,=\,{n\choose4}\;\text{ to }\;a_{n+1} \,=\,{n+1\choose4}\)

\(\displaystyle \text{To go from }\,\frac{n!}{4!\,(n-4)!}\;\text{ to }\;\frac{(n+1)!}{4!\,(n-3)!}\;\hdots\text{ multiply by }\,\frac{n+1}{n-3}\)


\(\displaystyle \text{Therefore: }\;a_{n+1} \;=\;\left(\frac{n+1}{n-3}\right)a_n\)