Explanation: In how many ways can 6 things be divided between 2 people? I have a question in a book which says in how many ways can 6 different things be divided between 2 boys and (my understanding of) the explanation goes something along the lines of:
Items: 1 1 1 1 1 1
The first item can be distributed to either boy 1 or boy 2, i.e. 2 choices the second item could be distributed to either boy 1 or boy 2 therefore you get 2 choices again.
Therefore the number of choices are $2\times2\times2\times2\times2\times2$.
However if I thought of it in another way such as you have 6 items therefore boy 1 can get all 6 and boy 2 could get 0, you could organise it into pairs such that:
\begin{array}{c c}
boy 1 & boy 2 \\
0 & 6 \\
1 & 5 \\
2 & 4 \\
3 & 3 \\
4 & 2 \\
5 & 1 \\
6 & 0
\end{array}
Therefore there are 7 ways of distributing the items. Where am I wrong with my thoughts? I cannot see why my way of thinking is wrong.
 A: The second method does not take care of the fact that items are distinct.
Assume items are $A,B,C,D,E,F$
In the second method, you counted $(1,5)$ as just one.
In fact, it can be
$A~~(B,C,D,E,F)\\
B~~(A,C,D,E,F)\\
C~~(A,B,D,E,F)\\
D~~(A,B,C,E,F)\\
E~~(A,B,C,D,F)\\
F~~(A,B,C,D,E)$
i.e., this (1,5) alone can happen in 6 different ways.
Each of the other cases can be considered in a similar way. Note that your count 7 is true if things are identical.
A: Your first method correctly counts the number of ways of distributing six different things to the two boys.
Your second method is wrong because it does not take account of the fact that the items are different. 
Your second method (if you do it correctly) would give the number of ways of dividing six identical things between two people. (I think with the second method you meant to count the 7 possible cases: Boy 1 gets $0,1,2,3,4,5$ or $6$ things.)
When the objects are distinguishable from each other, there are more distinguishable ways to distribute them. Suppose the objects are balls. If the balls are identical then there is only one way to give the first boy one ball (the balls cannot be distinguished from each other). If the balls are all different colours, then there are multiple ways of giving the first boy one ball (the ball can be six different colours).
