Question

There are 6 boxes numbered 1,2,... 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

• A. 5
• B. 21
• C. 33
• D. 60

Hint:

Each valid colouring is decided by where the run of green boxes starts and where it stops. Count the possible runs.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The green balls must sit in a run of neighbouring boxes with no gap, and there must be at least one green ball. Every other box gets a red ball. So each way of filling the boxes is fixed by choosing which block of consecutive boxes holds the green balls.

Step 2: Key Formula or Approach:
Count the blocks by their length k. A block of length k inside 6 boxes can start at box 1, 2, ... up to box \( 6 - k + 1 \), so there are \( 7 - k \) such blocks.

Step 3: Count for each length.
Length 1: 6 blocks (any single box).
Length 2: 5 blocks (1-2, 2-3, 3-4, 4-5, 5-6).
Length 3: 4 blocks.
Length 4: 3 blocks.
Length 5: 2 blocks.
Length 6: 1 block (all boxes green).

Step 4: Add them up.
\[ 6 + 5 + 4 + 3 + 2 + 1 = 21 \] Each block gives exactly one valid colouring, because all boxes outside the block must be red.

Step 5: Check the options.
Option 1 (5) counts only the blocks of length 2 and ignores all other lengths. Option 3 (33) and option 4 (60) are bigger than the count of blocks. There are only 21 runs, so 21 is right.

Final Answer:
There are 21 ways, option 2. \[ \boxed{21} \]