1. ## Probability combination!

Q In how many ways can n things be shared between 2 people.

And it used
(1+1)^n
But i dont get why they use (1+1)^n and im confused with the idea of 'sharing'.

P.s: how do i solve if it's between 3 people?

2. ## Re: Probability combination!

Originally Posted by gahyunkk
Q In how many ways can n things be shared between 2 people.

This is incorrect! Call the two people "A" and "B". Suppose n= 1. The one thing can be given to either A or B. There are 2= 2^1, not 2^1- 2= -1, ways to "share". Suppose n= 2. We can give both things to A or give both things to B or give the first to A and the second to B or give the first to B and the second to A. There are 4= 2^2, not 2^2= 2, different ways. In general, there are 2^n, not 2^n- 2, ways to distribute n things among 2 people.

And it used
(1+1)^n
But i dont get why they use (1+1)^n and im confused with the idea of 'sharing'.
I don't know why they wrote "1+ 1" but that is simply 2^n because there are 2 people.

P.s: how do i solve if it's between 3 people?
Pretty much the same way. If there are 3 people, each thing can be given to any one of the 3: there are 3^n ways to do that.

3. ## Re: Probability combination!

What if we suppose sharing has to be like two people both must have at least one(=both can't have zero)?

4. ## Re: Probability combination!

Then it's like you say, but you didn't assume that.

5. ## Re: Probability combination!

Originally Posted by gahyunkk
Q In how many ways can n things be shared between 2 people.

And it used
(1+1)^n
But i dont get why they use (1+1)^n and im confused with the idea of 'sharing'.

P.s: how do i solve if it's between 3 people?
I do not like this problem because the English word "share" is ambiguous. It is debatable whether n to person A and 0 to person B is sharing. In fact it could be construed that if n is even then there is only one way to share, in that A and B each get half, and if n is odd then there are only two ways to share, A gets (n - 1)/2 or (n + 1)/2. Moreover, it makes a difference whether the n objects are distinguishable. What seems to be meant by this problem is how many ways can n distinguishable objects be allocated between 2 people so that each gets something, given n is greater than 0.

The answer to that question is

$\displaystyle \sum_{i=1}^{n-1}\dbinom{n}{i} = \left(\sum_{i=0}^n\dbinom{n}{i}\right) - 1 - 1 = (1 + 1)^n - 2 = 2^n - 2.$

The expansion of (1 + 1)^n gives the binomial coefficients.

Ohh thanks !