Multiple choice

Given two three bit numbers a2a1a0 and b2b1b0 and c, the carry in, the function, that represents the carry generate function when these two numbers are added is

  1. $a_2b_2 + a_2a_1b_1 + a_2a_1a_0b_0 + a_2a_0b_1b_0 + a_1b_2b_1 + a_1a_0b_2b_0 + a_0b_2b_1b_0$
  2. $a_2b_2 + a_2b_1b_0 + a_2a_1b_1b_0 + a_1a_0b_2b_1 + a_1a_0b_2 + a_1a_0b_2b_0 + a_2a_0b_1b_0$
  3. $a_2 + b_2 + (a_2 \oplus b_2) + (a_1+ b_1 + (a_1\oplus b_1)+ (a_0 + b_0))$
  4. $a_2b_2 + \bar a_2 a_1b_1 + \overline{a_2a_1} a_0b_0 + \bar a_2 a_0 \bar b_1 b_0 + a_1\bar b_2b_1 + \bar a_1a_0\bar b_2b_0+a_0\overline{b_2b_1}b_0$
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A Correct answer
Explanation

For carry look ahead adder we know carry generate function— Where

As we are having two 3 bits number to add so final carry out will be C3- Putting value of Pi,Gi in 3

$C_3=(A_2.B_2)+(A_1.B_1)(A_2+B_2)+(A_0.B_0)(A_1+B_1)(A_2+B_2) (\text{TAKING }C0=0) $ $C_3=A_2.B_2 +A_1A_2B_1+A_1B_2B_1+(A_0B_0)(A_1A_2+A_1B_2+B_1A_2+B_1B_2) $ $C_3=A_2B_2+A_1A_2B_1+A_1B_2B_1+A_0A_1A_2B_0+A_0A_1B_0B_2+A_0A_2B_1B_0+A_0B_0B_1B_2 $

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