Consider a combination of T and D flip-flops connected as shown below. The output of the D flip-flop is connected to the input of the T flip-flop and the output of the T flip-flop is connected to the input of the D flip-flop. Initially, both $Q_0$ and $Q_1$ are set to 1 (before the $1^{\text{st}}$ clock cycle). The outputs

From the circuit diagram:

• The T flip-flop has input $T = Q_0$ and output $Q_1$.

• The D flip-flop has input $D = Q_1$ and output $Q_0$.

The next-state equations are:

Tracing cycle-by-cycle starting from initial state $Q_1 Q_0 = 11$:

• Initial (Cycle 0):$Q_1 = 1, Q_0 = 1$

• Cycle 1:

  $$Q_1(1) = 1 \oplus 1 = 0, \quad Q_0(1) = 1 \implies Q_1 Q_0 = 01$$

• Cycle 2:

  $$Q_1(2) = 1 \oplus 0 = 1, \quad Q_0(2) = 0 \implies Q_1 Q_0 = 10$$

• Cycle 3:

  $$Q_1(3) = 0 \oplus 1 = 1, \quad Q_0(3) = 1 \implies Q_1 Q_0 = 11$$

• Cycle 4:

  $$Q_1(4) = 1 \oplus 1 = 0, \quad Q_0(4) = 1 \implies Q_1 Q_0 = 01$$

Thus, $Q_1 Q_0$ after the $3^{\text{rd}}$ cycle is 11 and after the $4^{\text{th}}$ cycle is 01.

Hence, the correct option is B.