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.

2. ${\binom {n} {3}}=\frac{{n!}}{{\left( {3!} \right)\left( {n - 3} \right)!}} = \frac{{\left( n \right)\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{\left( {3!} \right)\left( {n - 3} \right)!}} = \frac{{\left( n \right)\left( {n - 1} \right)\left( {n - 2} \right)}}{{\left( 6 \right)}}$

3. thanks, but it has to include a basis, induction hypothesis, and induction step for it be a complete proof.

4. Originally Posted by jas05s
thanks, but it has to include a basis, induction hypothesis, and induction step for it be a complete proof.
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.

5. Then base case: For $n = 3....$ verify. Inductive step: For some integer $k$, statement holds true.

Prove that $P(k) \implies P(k+1)$.