Question

In a locality, there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are next to each other?

• A. 56
• B. 73
• C. 80
• D. 120
• E. None of the above

Hint:

In “non-adjacent selection” problems, convert them into combinations by reducing the total number of slots. The formula is \(\binom{n-k+1}{k}\). This saves time compared to manually listing cases.

Answer 1

The Correct Option is A

Step 1: Understand the problem.
There are \(10\) houses in a row. The thief wants to rob exactly \(3\) houses, but with the restriction that no two robbed houses can be \emph{adjacent}. This means whenever one house is chosen, its immediate left and right neighbors cannot be chosen.

Step 2: Standard combinatorial transformation.
When a restriction like “no two are adjacent” is applied, we can use the “gap method.” - Imagine first placing \(3\) robbed houses. - To ensure they are not adjacent, we place a gap (at least one unrobbed house) between them.

Step 3: Reformulate problem.
Suppose we mark the robbed houses as \(R\) and the unrobbed as \(U\). We want an arrangement like: \[ U \; R \; U \; R \; U \; R \; U \] Here, the extra \(U\)'s at ends and in between guarantee that no two \(R\)'s are next to each other.

Step 4: Formula approach.
The general formula is: \[ \binom{n-k+1}{k} \] where \(n = 10\) total positions (houses) and \(k = 3\) selected positions (robbed houses). This works because if \(k\) houses are chosen such that no two are adjacent, it is equivalent to choosing \(k\) positions from \((n-k+1)\) available slots after accounting for the gaps.

Step 5: Apply values.
\[ \binom{10-3+1}{3} = \binom{8}{3} \] \[ = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \] \[ = 56 \]

Step 6: Verify logic.
- If thief robs house 1, the next possible houses are from 3 to 10. - If thief robs house 2, then he cannot rob house 1 and 3, leaving other options. - Counting systematically matches the formula’s result.

Final Answer:
\[ \boxed{56} \]