Number of k-combinations of an n-set using orbit-stabilizer theorem

I have two similar questions:

Justify that a set of n elements has $$\binom nk$$ subsets of k elements by using the orbit-stabilizer theorem.

The orbit-stabilizer theorem says that $$|Orb(x)|=[G:Stab(x)]=|G||Stab(x)|$$

My attempt is if $$X = \{1,2,3,...,n \}$$ and S be the set of all subsets of X with k elements, then $$S_{n}$$ acts on $$S$$ with $$n!$$ permutations.

The orbit of $$\{1,2,3, ...,k \}$$ is S.

The order of the stabilizer should be $$k!(n-k)!$$(I do not know how to prove that!).

The second question:

For fi nite subgroups $$H$$ and $$K$$ of a group G, show that : $$|HK| = \frac{|H||K|}{|H\cap K|}$$ using the orbit-stabilizer theorem.

My attempt: with the direct product $$H \times K$$, we have $$(h,k)*g := hgk^{-1}$$.

Any idea on whether it is right or wrong.

• The second question is covered at math.stackexchange.com/questions/168942/… – Arnaud D. Nov 7 '18 at 10:41
• Oh ... Many thanks – Nawal Nov 7 '18 at 10:43
• As for the first one, to find the order of the stabilizer you can try to prove that it is isomorphic to $S_k\times S_{n-k}$. – Arnaud D. Nov 7 '18 at 10:48