One Munnabhai gets admitted to a psychiatric hospital. In his room, there are two bulbs connected to two different switches, independently. One night, while accompanying him, his buddy Circuit notices that there is, initially, no light in the room, but whenever a mosquito bites him he switches on the light and then immediately switches it off. Throughout the night, Munnabhai presses the switch(on/off) six times. Finally when Munnabhai stops playing around with the switches, Circuit notices that there is no light in the room. In how many ways Munnabhai ends up having no light in his room by pressing the given switches on or off exactly six times?

Options

A. A) 3
B. B) 4
C. C) 6
D. D) 8

Answer

To solve the problem, we need to determine how many ways Munnabhai can press the switches such that after six presses, both bulbs remain off.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- There are two bulbs and two switches, and initially, both bulbs are off.
- Munnabhai presses the switches a total of six times.

2. **Switch Behavior**:
- Each press of a switch toggles the state of the corresponding bulb (on to off or off to on).
- Since there are two bulbs, each press can either toggle bulb 1 or bulb 2.

3. **Total Presses**:
- Munnabhai presses the switches six times. Each press has two independent choices (bulb 1 or bulb 2).

4. **Calculating Total Combinations**:
- For each of the six presses, there are 2 choices (either press switch for bulb 1 or switch for bulb 2).
- Therefore, the total number of ways to press the switches is calculated as:
\[ \text{Total Ways} = 2^6 = 64 \]

5. **Condition for No Light**:
- For the bulbs to remain off after six presses, the number of times each bulb is toggled must be even (0, 2, 4, or 6 times).
- This is because toggling a bulb an even number of times will leave it in the off state.

6. **Distribution of Presses**:
- Let \( x \) be the number of times bulb 1 is toggled, and \( y \) be the number of times bulb 2 is toggled.
- We need to find the combinations of \( x \) and \( y \) such that \( x + y = 6 \) and both \( x \) and \( y \) are even.

7. **Possible Values**:
- The possible even values for \( x \) and \( y \) that sum to 6 are:
- \( (0, 6) \)
- \( (2, 4) \)
- \( (4, 2) \)
- \( (6, 0) \)

8. **Calculating Combinations**:
- For each pair \( (x, y) \), we can calculate the number of ways to arrange the presses:
- For \( (0, 6) \): 1 way (all presses for bulb 2)
- For \( (2, 4) \): \( \frac{6!}{2!4!} = 15 \) ways
- For \( (4, 2) \): \( \frac{6!}{4!2!} = 15 \) ways
- For \( (6, 0) \): 1 way (all presses for bulb 1)

9. **Total Combinations for No Light**:
- Total ways = \( 1 + 15 + 15 + 1 = 32 \)

### Final Answer:
Thus, the total number of ways Munnabhai can press the switches such that there is no light in the room after six presses is **32**.