In how many ways can $n$ number of students be divided into two teams such that each team has at least one student.

This is what I did:
Let $x_1$ be the number of students in the one team and $x_2$ be the number of students in the other team.Then number of ways of dividing students is equivalent to

$x_1+x_2=n$ such that $x_1,x_2>=1$.
I let $y_1=x_1-1$ and $y_2=x_2-1$.
Then the number of ways of dividing students are $n-1 \choose 1 $=$n-1$

The given answer in the book is $2^{n-1}-1$.
How is this answer obtained.Can someone please show me what's wrong with what I did.

  • $\begingroup$ Possible duplicate of this one $\endgroup$ – dajoker Oct 24 '14 at 14:18

Let's first say that there is a team $A$ and a team $B$. For each of the $n$ student there is a decision to make: in $A$ or in $B$. This gives at first $2^n$ possibilities but taking into account that both teams are not allowed to be 'empty' $2$ of these possibilities must be dropped. Then we have $2^n-2$ possibilities left. Secondly the teams are not distinguishable wich means that every possibility has in fact been counted twice. We repair this by dividing by $2$ and end up with $2^{n-1}-1$ possibilities.

  • $\begingroup$ I understand your answer but I can't see why my approach is wrong.What have I calculated? $\endgroup$ – clarkson Oct 24 '14 at 15:15
  • 1
    $\begingroup$ In your calculation only the quantities of the teams play a part. Secondly if there are e.g. $3$ students then the outcomes $(x_1,x_2)=(1,2)$ and $(x_1,x_2)=(2,1)$ are (wrongly) looked at as different. The point here is that the students are distinguishable and the teams are not. $\endgroup$ – drhab Oct 24 '14 at 15:26

One of the students is Alphonse. Let $N$ be the set of non-Alphonses. We need to choose a subset of $N$ to be on Alphonse's team, but we cannot choose all of $N$. The set $N$ has $2^{n-1}-1$ subsets other than $N$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.