Product of product of all subset of size k of a set?

Question

example given set={$1,2,3$} size={$2$}

then ($1,2),(2,3),(1,3$) so product of each is $2,6,3$ and the product is $2*6*3=36$ again.

so i was looking for a method to do ?

Answer 1

Let's assume the set we're dealing with is finite. Let $S = \{s_1, s_2, \cdots, s_n \}$, and fix the size we're considering as $k$.

There are $\frac{n!}{k! (n-k)!}$ total subsets of size $k$. Now each element of $S$ will be a member of some (but not all) of these subsets. In particular, if we force $s_j$ to be in subset, there are $k-1$ remaining members to pick from the remaining $n-1$ members of $S$. Thus, there are exactly $\frac{(n-1)!}{(k-1)!(n-k)!}$ subsets of size $k$ containing $s_j$.

Since this holds for any $j$, and muliplication in commutative, we find the value you're l0oking for, $V$ is given by:

$$V = \prod_{j=1}^{n}s_j^r$$ where $r$ is given by: $$r = \frac{(n-1)!}{(k-1)!(n-k)!}$$

---

Applying this to your example above, we have $n =3$ and $k=2$, which yields $r = 2$. Then we have: $$\begin{align} V &= 1^2 *2^2 *3^2 \\ &= 1*4*9 \\ &=36 \end{align}$$

Answer 2

Consider the $k-$sets of $\{a_1,\cdots,a_n \}$ each element will occur $\binom{n-1}{k-1}$ times in a given set, so the product of the products of the sets will be $\color{red}{\left(\prod_{i=1}^{n} a_i \right)^{\binom{n-1}{k-1}}}$.