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Given two three bit numbers $a_{2}a_{1}a_{0}$ and $b_{2}b_{1}b_{0}$ and $c$ the carry in, the function that represents the carry generate function when these two numbers are added is: 

  1. $a_{2}b_{2}+a_{2}a_{1}b_{1}+a_{2}a_{1}a_{0}b_{0}+a_{2}a_{0}b_{1}b_{0}+a_{1}b_{2}b_{1}+a_{1}a_{0}b_{2}b_{0}+a_{0}b_{2}b_{1}b_{0}$
  2. $a_{2}b_{2}+a_{2}b_{1}b_{0}+a_{2}a_{1}b_{1}b_{0}+a_{1}a_{0}b_{2}b_{1}+a_{1}a_{0}b_{2}+a_{1}a_{0}b_{2}b_{0}+a_{2}a_{0}b_{1}b_{0}$
  3. $a_{2}+b_{2}+(a_{2}\oplus b_{2}) ( a_{1}+b_{1}+(a_{1}\oplus b_{1})+(a_{0}+b_{0}))$
  4. $a_{2}b_{2}+\overline{a_{2}}a_{1}b_{1}+\overline{a_{2}a_{1}}a_{0}b_{0}+\overline{a_{2}}a_{0}\overline{b_{1}}b_{0}+a_{1}\overline{b_{2}}b_{1}+\overline{a_{1}}a_{0}\overline{b_{2}}b_{0}+a_{0}\overline{b_{2}b_{1}}b_{0}$
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6 Answers

Best answer
77 votes
77 votes

$c_1=a_0 b_0$

$c_2 = a_1b_1 + a_1c_1 + b_1c_1$

$c_3=a_2b_2 + a_2c_2 + b_2c_2$
$\quad=a_2b_2 + a_2a_1b_1 + a_2a_1c_1 +a_2b_1c_1 + b_2a_1b_1 +b_2a_1c_1 + b_2b_1c_1$
$\quad= a_2b_2 + a_2a_1b_1 +a_2a_1a_0b_0 + a_2b_1a_0b_0 + b_2a_1b_1 + b_2a_1a_0b_0 +b_2b_1a_0b_0$

Option is A.

Considering the carry in function $c$, $c_1 = a_0b_0 + a_0c + b_0c$, but $c$ is missing in all options and hence ignored.

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44 votes
44 votes

(A) is correct option!

21 votes
21 votes

First we have to know how carry can be generated.

1) If we add 2-bits carry can only be generated if both digits are 1.

so c=a0b0


2) If we add two 2-bit numbers then carry can be generated in two cases ,

a)MSB of both number is 1 

b)One MSB is 0 and Other one is 1 and both a0 and b0 are =1

so C=a1b1 + (a1+b1)a0b0


3) If we add two 3-bit numbers then carry can be generated in 3 cases ,

a) MSB of both numbers is = 1

b) MSB of one of the digit is 1 and a1 and b1 both are 1.

c) MSB of one of the digit is 1 and other MSB=0 and one of a1 , b1 is 1 and other 0 but LSB of both digit must be 1

so C= a2b2 + (a2+b2)a1b1 + (a2+b2)(a1+b1)a0b0  

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