Last Diwali Chhota Bheem got 10 distinct coloured candles placed in a row. Then he asked his friend to lit 3 candles.
Find the number of ways of selection of candles so that all the at least 2 candles are consecutive.

Options

A. 66
B. 33
C. 17
D. 64

Answer

To solve the problem of selecting 3 candles such that at least 2 of them are consecutive, we can break it down into three cases:

### Step 1: Understand the total number of candles
We have a total of 10 distinct colored candles arranged in a row. We denote them as A, B, C, D, E, F, G, H, I, J.

### Step 2: Define the cases
We will consider three cases based on the arrangement of the selected candles:

1. **Case 1**: Two consecutive candles are at the ends (either A and B or I and J).
2. **Case 2**: Two consecutive candles are in the middle (like BC, CD, DE, etc.).
3. **Case 3**: All three candles are consecutive.

### Step 3: Calculate Case 1
In this case, we can have the following pairs of consecutive candles:
- (A, B)
- (I, J)

For each of these pairs, we can choose the third candle from the remaining candles that are not neighbors to the selected pair.

- For (A, B): The third candle can be any of C, D, E, F, G, H, I, J (total 8 options).
- For (I, J): The third candle can be any of A, B, C, D, E, F, G, H (total 8 options).

Thus, for Case 1, we have:
- Total ways = 2 (pairs) × 8 (choices for the third candle) = 16 ways.

### Step 4: Calculate Case 2
In this case, we can have the following pairs of consecutive candles:
- (B, C), (C, D), (D, E), (E, F), (F, G), (G, H), (H, I)

This gives us 7 pairs. For each pair, we need to select the third candle, which cannot be adjacent to the selected pair.

For example:
- For (B, C): The third candle can be A, D, E, F, G, H, I, J (total 8 options).
- For (C, D): The third candle can be A, B, E, F, G, H, I, J (total 8 options).
- Similarly for the other pairs.

However, we need to exclude the adjacent candles.

After calculation, we find that for each of the 7 pairs, we can select the third candle from 6 remaining options (excluding the two adjacent candles).

Thus, for Case 2, we have:
- Total ways = 7 (pairs) × 6 (choices for the third candle) = 42 ways.

### Step 5: Calculate Case 3
In this case, we select 3 consecutive candles. The possible selections are:
- (A, B, C), (B, C, D), (C, D, E), (D, E, F), (E, F, G), (F, G, H), (G, H, I), (H, I, J)

This gives us a total of 8 ways to select 3 consecutive candles.

### Step 6: Combine all cases
Now we sum the results from all three cases:
- Case 1: 16 ways
- Case 2: 42 ways
- Case 3: 8 ways

Total = 16 + 42 + 8 = 66 ways.

### Final Answer
The total number of ways to select 3 candles such that at least 2 are consecutive is **66**.

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