Three boxes are coloured red, blue and green and so are three balls. In how many ways can one put the balls one in eachbox such that no ball goes into the box of its own colour?

Correct option is B

Given:

• 3 boxes: Red, Blue, Green
• 3 balls: Red, Blue, Green
• Condition: No ball is placed in the box of the same color.

Concept: Derangement

We want the number of derangements of 3 distinct items — i.e., permutations where no element appears in its original position.

Solution:

The formula for number of derangements of 3 items:

​$$!3 = 3! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} \right) = 6 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} \right) = 6 \times \left( \frac{1}{3} \right) = 2$$​​

Final Answer: (B) 2

There are 2 ways to place the balls such that no ball goes into the box of its own color.