Subsets of a power set. How many subsets T of the power set of A contain at most 2 elements if the cardinality of A is n where n is a natural number ?
The number of elements in power set A = 2^n.  But i dont know how to work out the number of subsets of the power set that contain at most 2 elements.  Any answers and i will be grateful.
 A: If  $ Card (A)=n$ which  $n \in \mathbb N $ then the number of subsets of $A$ that have exactly $ j $ elements is $  \binom {n}{j}$
Then the answer of the questin is 
${1+ \binom{n}{1} + \binom{n}{2}}$= ${  \frac{n^2+n+2}{2}}$
Note : 1 is for empty set
A: Suppose $A=\{a_1,a_2,\dots,a_n\}$, then 
$$\mathcal P(A)=\{\emptyset,\overbrace{\{a_1\},\{a_2\},\dots,\{a_n\}}^{\binom{n}{1} \text{ sets}},\overbrace{\{a_1,a_2\},...}^{\binom{n}{2}\text{ sets}},\overbrace{\{a_1,a_2,a_3\}}^{\binom{n}{3} \text{ sets}},...,A\}.$$
Recall that $\binom{n}{k}$ tells you the number of sets you can make with $k$ elements from a set of $n$ elements.
Then if we add everything:
$$|\mathcal P(A)|=\sum\limits_{i=0}^n\binom{n}{i}=\sum\limits_{i=0}^n\binom{n}{i}(1^i)(1^{n-i})\stackrel{\text{by Binomial Theorem}}{=}(1+1)^n=2^n.$$
Why do we sum $\binom{n}{0}$ and $\binom{n}{n}$? Because of the number of sets with no elements (this is the empty set) and the number of sets with $n$ elements (this is $A$). Therefore you are interested in:
$$\sum\limits_{i=0}^2\binom{2^n}{i}.$$
