Prove that a set with n elements has n(n-1)(n-2)/6 subsets containing exactly three elements whenever n is an integer greater than or equal to 3.
Please help, I understand the basics but I am currently stuck on this problem.
Prove that a set with n elements has n(n-1)(n-2)/6 subsets containing exactly three elements whenever n is an integer greater than or equal to 3.
Please help, I understand the basics but I am currently stuck on this problem.
$\displaystyle {\binom {n} {3}}=\frac{{n!}}{{\left( {3!} \right)\left( {n - 3} \right)!}} = \frac{{\left( n \right)\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{\left( {3!} \right)\left( {n - 3} \right)!}} = \frac{{\left( n \right)\left( {n - 1} \right)\left( {n - 2} \right)}}{{\left( 6 \right)}}$
This is a prefect example of why many of us think that the current state of mathematics training is in the dumps. Given how many students have no clear idea what ‘induction proofs’ are all about, why complicate things with such a problem?
Is this a set theory class? If it is then I am wrong.
Otherwise, I stand by what I have written.