Let $A= \{1,2,3,4,5,6,7,8,9,0,20,30,40,50\}$. 1. How many subsets of size 2 are there? 2.How many subsets are there altogether? 
Let $A= \{1,2,3,4,5,6,7,8,9,0,20,30,40,50\}$.
  1. How many subsets of size $2$ are there? 
  2.How many subsets are there altogether?

Answer: 
1) I think there are $7$ subsets of size two are there, since $14$ elements$/2=7$ and also by grouping them into pairs it shows there are $7$ groups of pairs as follows; $\{\{1,2\}, \{3,4\}, \{5,6\}, \{7,8\}, \{9,0\}, \{20,30\}, \{40,50\}\}$
2) I choose $15$, but I'm not sure.
However, are my answers valid?  Hints are much appreciated.


How many subsets of size $2$ are there?

Set $A$ has $14$ elements.  In selecting a subset of size $2$, we have $14$ choices for the first element and $13$ choices for the second element.  Thus, at first glance, it appears we have $14 \cdot 13$ possible subsets.  However, the order in which the elements are selected does not matter.  Selecting the element $1$ and then the element $2$ results in the same subset as selecting the element $2$ and then the element $1$ since $\{1, 2\} = \{2, 1\}$.  Dividing by the $2! = 2$ orders in which two elements can be selected yields 
$$\frac{14 \cdot 13}{2}$$
subsets with two elements.  
In general, if we wish to form a subset of $k$ elements from a set of $n$ elements, we have $n$ choices for the first element, $n - 1$ choices for the second element, $n - 2$ choices for the third element, and so forth until we are left with $n - (k - 1) = n - k + 1$ choices for the $k$th element.  Hence, there are 
$$n(n - 1)(n - 2) \cdots (n - k + 1)$$
ways to make an ordered selection of $k$ elements.  However, the same $k$ elements in the subset can be selected in 
$$k(k - 1)(k - 2) \cdots 1 = k!$$
orders.  Thus, the number of subsets of size $k$ that we can select from a set with $n$ elements is 
$$\frac{n(n - 1)(n - 2) \cdots (n - k + 1)}{k(k - 1)(k - 2) \cdots 1} = \frac{n(n - 1)(n - 2) \cdots (n - k + 1)}{k!}$$
Multiplying the numerator and denominator by $(n - k)!$ yields 
$$\frac{n(n - 1)(n - 2) \cdots (n - k + 1)(n - k)!}{k!(n - k)!} = \frac{n!}{k!(n - k)!}$$
The number 
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
is the number of subsets of $k$ elements that can be selected from a set with $n$ elements.
Hence, the number of subsets of set $A$ is 
$$\binom{14}{2} = \frac{14!}{2!12!} = \frac{14 \cdot 13 \cdot 12!}{2 \cdot 1 \cdot 12!} = \frac{14 \cdot 13}{2} = 91$$
as Andre Nicolas and others have indicated.

How many subsets are there altogether?

A subset of a set $S$ with $n$ elements is determined by which elements it includes.  When forming a subset, we have two choices for each of the $n$ elements, to include the element in the subset or to not include it in the subset.  Hence, a set $S$ with $n$ elements has $2^n$ subsets.
Since set $A$ has $14$ elements, it has $2^{14}$ subsets. 
While it would be impractical to list all $2^{14} = 16,384$ subsets of set $A$, we can verify that the set $\{1, 2, 3\}$ has $2^3 = 8$ subsets.  They are 
$$\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}$$
