Two truth tellers and two liars are positioned in a line, where every person is distinguishable. How many ways are there to position these four people such that everyone claims that all people directly adjacent to them are liars?

$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }12\qquad\textbf{(E) }16$

1. Let the truth tellers be denoted as \( T_1 \) and \( T_2 \), and the liars be denoted as \( L_1 \) and \( L_2 \).
2. Since a liar claims that all people directly adjacent to them are liars, the people directly adjacent to a liar must actually be truth-tellers. Therefore, no liar can be next to another liar, and no truth-teller can be next to another truth-teller.
3. This implies that the arrangement must alternate between truth-tellers and liars. The possible alternating sequences are \( TLTL \) and \( LTLT \).
4. For the sequence \( TLTL \):
   - We can choose any of the 2 truth-tellers for the first \( T \) position, and any of the remaining 1 truth-teller for the second \( T \) position. This gives us \( 2! = 2 \) ways to arrange the truth-tellers.
   - Similarly, we can choose any of the 2 liars for the first \( L \) position, and any of the remaining 1 liar for the second \( L \) position. This gives us \( 2! = 2 \) ways to arrange the liars.
   - Therefore, the total number of ways to arrange \( TLTL \) is \( 2 \times 2 = 4 \).
5. For the sequence \( LTLT \):
   - We can choose any of the 2 liars for the first \( L \) position, and any of the remaining 1 liar for the second \( L \) position. This gives us \( 2! = 2 \) ways to arrange the liars.
   - Similarly, we can choose any of the 2 truth-tellers for the first \( T \) position, and any of the remaining 1 truth-teller for the second \( T \) position. This gives us \( 2! = 2 \) ways to arrange the truth-tellers.
   - Therefore, the total number of ways to arrange \( LTLT \) is \( 2 \times 2 = 4 \).
6. Adding the number of ways for both sequences, we get \( 4 + 4 = 8 \).

The final answer is \(\boxed{8}\).