0

$V=V_{1}\sqcup V_{2}$ we will understand that V is a sum of disjont subsets $V_{1}$ and $V_{2}$.If V={1} then $V_{1}=V$, $V_{2}=\emptyset$ and $V_{1}=\emptyset,V_{2}=V$ we have two possibility.If $V$={1,2}, then$V_{1}=V,V_{2}=\emptyset$ and $V_{1}$={1},$V_{2}$={2}and $V_{1}$={2},$V_{2}$={1} and $V_{1}=\emptyset$,$V_{2}=V$.We have 4 possibility.$$$$My question is if we take $V=${1,...,n} then how many possibilty occur and how it can be formulized ?

Asaf Karagila
  • 405,794
1ENİGMA1
  • 1,247

1 Answers1

0

We will have $2^n $ possibilities.

Indeed, whenever you have $V= \{1,2,...,n \}$, you are seting $V_1$ to be a subset of $V$ , and $V_2$ to be its complement. Hence how many subsets of $ \{1,2,...,n \}$ we have ( including $\phi$)? The cardinality is $2^n$.

Nizar
  • 2,810
  • Thank you Nizar.Can you say any book name for this proof .If we take $V=V_{1}\sqcup...\sqcup V_{n}$ and $V_{i} \ne \emptyset$ i.e. not including $\emptyset$ then how it can be formulized ? – 1ENİGMA1 Nov 03 '15 at 14:39
  • Sorry I dont have a specific book for the proof I wrote. But for the problem you stated, I think you It can fomulated like "having n different balls, how can you distribute them in k boxes , so that non of the boxes is empty" right ? for this you can see http://math.stackexchange.com/questions/58753/unique-ways-to-keep-n-balls-into-k-boxes – Nizar Nov 03 '15 at 14:58
  • Thank you very much Nizar.How we can apply this for example $V=${1,2} then $V_{1}=${1},$V_{2}=${2} and $V_{1}$={2},$V_{2}=${1} we have two possibility. $\binom{n-1}{k-1}$ . – 1ENİGMA1 Nov 03 '15 at 15:20
  • Note that $ \dbinom{n-1}{k-1}$ is the answer for the case where all the $n$ given balls are identical. However here in your case the numbers are not identical (what I mean is that your set is ${1,2 } $ and not ${ 1,1}$). Try to generalize the result to be compatible with your case. I will try also thinking for the solution. The above comment was just a hint . – Nizar Nov 03 '15 at 15:27