1. ## ordered pair!!

Determine the number of ordered pairs (A,B), where A is a subset of B and B is a subset of {1,2,....,n}.

2. Originally Posted by vivian
Determine the number of ordered pairs (A,B), where A is a subset of B and B is a subset of {1,2,....,n}.
For a particular subset $B \subseteq \left\{ {1,2,3, \cdots ,n} \right\}$ and B has k elements then there are $2^k$ subsets of B. Hence there are $2^k$ ordered pairs (A,B) where $A \subseteq B$. BUT that is only for one particular subset.

Let’s do the general case: for any number j, $1 \le j \le n$, there are ${n \choose j}$ subsets of $\left\{ {1,2,3, \cdots ,n} \right\}$ having j elements. Thus there are ${n \choose j}2^j$ pairs of the form (A,B) where $A \subseteq B$ and |B|=j.

The total count: $\sum\limits_{j = 0}^n {{n \choose j} 2^j }$.

3. Originally Posted by Plato
The total count: $\sum\limits_{j = 0}^n {{n \choose j} 2^j }$.
Comparing that with the binomial formula $(x+y)^n = \sum\limits_{j = 0}^n {{n \choose j} x^jy^{n-j} }$, you see that the total count is $3^n$ (put x=2, y=1).

You can see this answer directly from the original question if you argue as follows. For each integer from 1 to n, there are three possibilities: (a) it belongs to both A and B, (b) it belongs to B but not A, (c) it belongs to neither. Total number of possible choices: $3^n$ (but I wouldn't have thought of that if I hadn't seen Plato's answer first).