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Let's analyze the logic gates used in the given combinatorial circuit.

The equation formed is \( X + \overline{Y}Z \). The circuit contains the following components:

1. **NOT gate**: This is used to negate \( Y \), giving \( \overline{Y} \).
2. **AND gate**: This takes \( \overline{Y} \) and \( Z \) as inputs, resulting in \( \overline{Y}Z \).
3. **OR gate**: This takes \( X \) and \( \overline{Y}Z \) as inputs, resulting in the final output \( X + \overline{Y}Z \).

Now, let's count the number of gates used:
1. One **NOT** gate for \( \overline{Y} \).
2. One **AND** gate for \( \overline{Y} \) and \( Z \).
3. One **OR** gate for the final output.

Hence, the total number of gates is **3**, not 4.

Thus, the statement "Number of logic gates used in the given combinatorial circuit are 4" is **False**.

Would you like more details or have any further questions?

### Related Questions:
1. What is the function of a NOT gate in a combinatorial circuit?
2. How do AND and OR gates combine in a circuit to create complex logic expressions?
3. What is the significance of the \( X + \overline{Y}Z \) expression in Boolean algebra?
4. How would the circuit change if the logic equation was \( X \overline{+ Y}Z \)?
5. Can you explain how truth tables relate to this type of logic circuit?

**Tip:** When counting gates in a circuit, always consider the basic components (NOT, AND, OR, etc.) used to implement the Boolean expression.

Number of logic gates used in the given combinatorial circuit are 4. True or False?

Explore the logic gate counting in a combinatorial circuit with the Boolean expression X + \overline{Y}Z. This problem helps in understanding the number and types of gates used in digital circuits, including NOT, AND, and OR gates.

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