Question

There are six boxes numbered 1, 2, 3, 4, 5, 6. Each box is to be filled up either with a white ball or a black ball in such a manner that at least one box contains a black ball and all the boxes containing black balls are consecutively numbered. The total number of ways in which this can be done equals:

• A. 15
• B. 21
• C. 63
• D. 64

Hint:

When elements must be "all consecutive," count non-empty intervals. For \(n\) positions, the number of intervals is \(\frac{n(n+1)}{2}\).

Answer 1

The Correct Option is B

Step 1: Model the condition. 
"All black boxes are consecutive" \(\Rightarrow\) the set of black boxes must form a single contiguous block (interval) among the 6 positions. 

Step 2: Count all non-empty intervals among 6 positions. 
Choose the start and end of the black block: for length \(1\) there are \(6\) choices; for length \(2\), \(5\) choices; \(\dots\); for length \(6\), \(1\) choice. 
\[ \text{Total ways} = 6+5+4+3+2+1 = \frac{6\cdot 7}{2} = 21. \] \[ \boxed{21} \]

Answer 2

This count can also be found by viewing the arrangement as a composition: some white boxes, then a block of black boxes, then more white boxes.

Setting up the composition. Write the arrangement as \(W^aB^kW^b\), meaning \(a\) white boxes, then \(k\ge1\) consecutive black boxes, then \(b\) white boxes, with \[ a+k+b=6,\quad a,b\ge0,\quad k\ge1. \]

Counting for each block length \(k\). For a fixed \(k\) (from 1 to 6), \(a+b=6-k\), and the number of non-negative integer pairs \((a,b)\) satisfying this is \(6-k+1=7-k\). Summing over all possible \(k\): \[ \sum_{k=1}^{6}(7-k)=6+5+4+3+2+1=21. \]

1. Option 15: This equals \(\binom{6}{2}\), which undercounts — it misses several valid placements of the black block; ruled out.
2. Option 21: Matches the composition count exactly.
3. Option 63: Would count arrangements where black boxes don't need to be consecutive, which is not the condition given; ruled out.
4. Option 64: This is \(2^6\), the count of ALL black/white arrangements including the all-white case and non-consecutive ones — far too permissive; ruled out.

The block-composition method confirms the same total of 21 valid ways.

So the correct answer is 21.