The truth table for the given logic circuit is: (Inputs $A, B$ and Output $E$)

The Boolean expression $Y = A \bar{B} C + \bar{A} \bar{C}$ can be realized with which of the following gate configurations?
$A.$ One $3$-input $\text{AND}$ gate,$2$ $\text{NOT}$ gates,one $2$-input $\text{AND}$ gate,and one $2$-input $\text{OR}$ gate.
$B.$ One $3$-input $\text{AND}$ gate,$2$ $\text{NOT}$ gates,one $2$-input $\text{NAND}$ gate,and one $2$-input $\text{OR}$ gate.
$C.$ One $3$-input $\text{OR}$ gate,$3$ $\text{NOT}$ gates,and one $2$-input $\text{AND}$ gate.
Choose the correct answer from the options given below.

The output of a $NAND$ gate is shown in the truth-table ($A$ and $B$ are inputs, $Y$ is output). Identify the correct truth-table.

Which logic gate is represented by the following truth table?

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

What will be the outputs $Y$ for the inputs $(A = 1, B = 0)$,$(A = 1, B = 1)$,and $(A = 0, B = 0)$ respectively?

In the following combination of logic gates,the output $Y$ can be written in terms of inputs $A$ and $B$ as: