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

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- Jan 6th 2008, 01:07 PMvivianordered 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}.

- Jan 6th 2008, 01:43 PMPlato
For a particular subset $\displaystyle B \subseteq \left\{ {1,2,3, \cdots ,n} \right\}$ and

*B*has*k*elements then there are $\displaystyle 2^k$ subsets of*B*. Hence there are $\displaystyle 2^k$ ordered pairs*(A,B)*where $\displaystyle A \subseteq B$. BUT that is only for one particular subset.

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

The total count: $\displaystyle \sum\limits_{j = 0}^n {{n \choose j} 2^j } $. - Jan 7th 2008, 12:00 AMOpalg
Comparing that with the binomial formula $\displaystyle (x+y)^n = \sum\limits_{j = 0}^n {{n \choose j} x^jy^{n-j} } $, you see that the total count is $\displaystyle 3^n$ (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: $\displaystyle 3^n$ (but I wouldn't have thought of that if I hadn't seen**Plato**'s answer first).