Since the exact circuit diagram is missing from the text, we deduce its logic from the standard NEET 2016 question it refers to. The circuit consists of an AND gate taking inputs A and B, whose output is fed into a NAND gate along with the third input C. The Boolean expression for this circuit is $Y = \overline{(A \cdot B) \cdot C}$.

Case 1: When A = 0, B = 0, and C = 0. Output of the AND gate = $0 \cdot 0 = 0$. Inputs to the NAND gate are 0 and 0. Output $Y = \overline{0 \cdot 0} = \overline{0} = 1$.

Case 2: When A = 1, B = 1, and C = 1. Output of the AND gate = $1 \cdot 1 = 1$. Inputs to the NAND gate are 1 and 1. Output $Y = \overline{1 \cdot 1} = \overline{1} = 0$.

Therefore, the sequence of outputs is 1, 0.