SOLUTION: For a set of two elements find the number of different subsets of 1 element. Use row 2 of pascals triangle to find the answer.

It's just like question 1146008 that I answered so I'll just copy and paste
from it.  That question there was:

"suppose 5 fair coins are tossed. use pascals triangle to find the number of
ways obtaining exactty 4 heads."

A coin has two "elements", a 'head' and a 'tail'. 

Remember the rule I gave you: 

Suppose N fair coins are tossed. To find the number of ways of
obtaining exactly R heads, go down to the (N+1)st row, and then
go right to the (R+1)st number.

Your problem here is 

Suppose 2 fair coins are tossed. To find the number of ways of
obtaining exactly R heads, go down to the 3rd row, and then
go right to the 2nd number.

But wait! The difficulty here is the fact that your teacher and book
leaves off the top "1" from Pascal's triangle (which makes it
not be a "triangle" but a trapezoid!!!) but it changes the rule 
from the (N+1)st row to the Nth row.  But it doesn't change the 
rule for the (R+1)st number in that row. [Your teacher should 
call it "Pascal's trapezoid" instead of "Pascal's triangle". lol]

So your Pascal's "triangle" looks like this trapezoid:

                    1     1
                 1     2     1
              1     3     3     1
           1     4     6     4     1

with the top "1" lopped off.  So your rule is

Suppose N fair coins are tossed. To find the number of ways of
obtaining exactly R heads, go down to the Nth row, and then
go right to the (R+1)st number.

You will use N as 2 and R as 1.  So it will be the same as

Suppose 2 fair coins are tossed. To find the number of ways of
obtaining exactly R heads, go down to the 2nd row, and then
go right to the 2nd number.  I made that number red above.

Edwin