GATE IT 2005 | Question: 46

Question

A line $L$ in a circuit is said to have a $stuck-at-0$ fault if the line permanently has a logic value $0$. Similarly a line $L$ in a circuit is said to have a $stuck-at-1$ fault if the line permanently has a logic value $1$. A circuit is said to have a multiple $stuck-at$ fault if one or more lines have stuck at faults. The total number of distinct multiple $stuck-at$ faults possible in a circuit with $N$ lines is 

• A. $3^N$
• B. $3^N - 1$
• C. $2^N - 1$
• D. $2$

Comments

https://www.geeksforgeeks.org/gate-gate-it-2005-question-46/

Answer 1

Answer should be ${3^N}-{1}$.

Explanation:

The total possible combinations (i.e., a line may either be at fault (in $2$ ways i.e stuck at fault $0$ or $1$) or it may not be, so there are only $3$ possibilities for a line ) is ${3^N}$.

In only one combination the circuit will have all lines to be correct (i.e not at fault.)

Hence ${3^N}-{1}$ (as it has been said that circuit is said to have multiple stuck up fault if one or more line is at fault)

Comments

If anyone of the line is stuck-at 0 or 1, then it doesn’t matter whether other lines are stuck-at or not. For each line we will have 3 cases: 

1. stuck-at-0

2. stuck-at-1

3. Working fine

Hence $3^n$ such cases, but this also counts the case where all the lines are working fine. So subtracting it $3^n – 1$

---

This will solve your doubt @Pranavpurkar

Gate 2005 | A line L in a circuit is said to have a stuck-at-0 fault if the line permanently has a - YouTube

---

Abhrajyoti00

Thanks:)

Answer 2

In a line, there can be three possibilities :
1. Stuck-at 0 fault
2. Stuck-at 1 fault
3. No fault

Thus, total combinations = 3^N

It is mentioned that one or more lines have stuck at faults.
So, a case in which there is no fault in any line i.e. all lines are correct can not occur.
Total combinations = 3^N – 1

 
Thus, option (B) is correct.