0

Given a set $A$ which contains $n$ elements. For any two distinct subsets $A_{1}$, $A_{2}$ of the given set $A$, we fix the number of elements of $A_1 \cap A_2$. Find the sum of all the numbers obtained in the described way.

My solution was $$\sum_{i=1}^{n} {i}\times \binom{n}{i}\times \left( {2^{n-1}-1} \right)$$ I don't know if this is the right solution and if it is, can it be simplified more or is it ok in this form?

  • I solved this problem, but am new to latex so i cant properly write it up. Will try to add it as soon as i figure it out. Just I am not sure in my solution. –  Apr 01 '17 at 20:45
  • If you have trouble with MathJax, this question might help you : https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Arnaud D. Apr 01 '17 at 21:01
  • What do you mean by "fix the number of..."? – DanielWainfleet Apr 02 '17 at 03:42

1 Answers1

0

Your answer is close but the factor of $2^{n - 1} - 1$ is not correct. (As an aside, your sum as written can be simplified using the same methods as below.)

Let $i = |A_1 \cap A_2|$. We would like to find the number of ways to construct $A_1$ and $A_2$ for each integer $1 \le i \le n - 1$ (note that if $i = n$, then $A_1 = A_2$, so there does not exist any two such distinct subsets). There are $\binom{n}{i}$ ways to choose $i$ common elements for $A_1$ and $A_2$. For each of the $n - i$ remaining elements of $A$, there are three places we can place each element: $A_1$, $A_2$, or neither. Since we cannot place all elements in neither set (otherwise $A_1 = A_2$), there are $\frac{1}{2}(3^{n - i} - 1)$ ways to place the remaining elements, where the factor of $1/2$ comes from the fact that we do not care about the order of $A_1$ and $A_2$. Therefore, there are $\binom{n}{i} \cdot \frac{1}{2} (3^{n - i} - 1)$ ways to select $A_1$ and $A_2$ given $i$, so the desired sum is

\begin{align} S &= \sum_{i = 1}^{n - 1} i \cdot \binom{n}{i} \cdot \frac{1}{2}\left(3^{n - i} - 1 \right) \\ &= \sum_{i = 1}^{n} i \cdot \binom{n}{i} \cdot \frac{1}{2}\left(3^{n - i} - 1 \right) \\ &= \frac{1}{2} \sum_{i = 1}^{n} i \binom{n}{i} 3^{n - i} - \frac{1}{2} \sum_{i = 1}^{n} i \binom{n}{i}. \end{align}

We now make use of the following identities (for a proof sketch, see here for example):

$$ \sum_{i = 1}^{n} i \binom{n}{i} 3^{n - i} = n 4^{n - 1}; \qquad \sum_{i = 1}^{n} i \binom{n}{i} = n 2^{n - 1}. $$

Therefore,

\begin{align} S &= \frac{1}{2} \cdot n 4^{n - 1} - \frac{1}{2} \cdot n 2^{n - 1} \\ &= \frac{n}{2} \left(4^{n - 1} - 2^{n - 1} \right) = n\left(2^{2n - 3} - 2^{n - 2} \right). \end{align}

Alex B.
  • 545