Wierd Question on Subets

I will number these from right to left, starting with 1, going up to 5. The rightmost one is 1, the leftmost is 5. 

2 1

2 3

2 4

2 5 

2 1 3 

2 1 4 

2 1 5 

2 3 4 

2 3 5 

2 4 5

2 1 3 4 

2 1 3 5 

2 1 4 5 

2 3 4 5 

2 1 3 4 5 

Thus, there is a total of $\color{brown}\boxed{15}$ subsets with the ₯ symbol. 

Here's a better way to get the same answer. 

For groups of 2 including  ₯, there is $4 \choose 1$ choices, because 1 of them is already taken. 

For groups of 3 including  ₯, there is $4 \choose 2$ choices, because 1 of them is already taken. 

For groups of 4 including  ₯, there is $4 \choose 3$ choices, because 1 of them is already taken. 

For groups of 5 including  $₯$, there is $4 \choose 4$ choices, because 1 of them is already taken. 

Adding everything up, there is $\color{brown}\boxed{15}$ subsets that contain ₯.

Thank you guys!!

Yes, subsets of just 1 term are possible. !!