Three boxes are coloured red, blue and green and so are three balls. In how many ways can one put the balls one in each box such that no ball goes into the box of its own colour?
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Understanding the Problem
The problem asks us to find the number of ways to place three coloured balls (red, blue, and green) into three coloured boxes (red, blue, and green), such that no ball is placed in the box that matches its own colour. This is a classic problem in combinatorics related to derangements.
Total Possible Arrangements
First, let's consider the total number of ways to place the three distinct balls into the three distinct boxes without any restrictions. We can place the first ball into any of the 3 boxes, the second ball into any of the remaining 2 boxes, and the third ball into the last remaining box. The total number of arrangements (permutations) is:
\(3! = 3 \times 2 \times 1 = 6\)
Let's list these 6 total arrangements. Assume the boxes are in the order Red Box, Blue Box, Green Box. The arrangement shows which ball goes into which box in that order.
| Arrangement # | Red Box | Blue Box | Green Box | Matches Own Colour? |
|---|---|---|---|---|
| 1 | Red Ball | Blue Ball | Green Ball | All match (Red, Blue, Green) |
| 2 | Red Ball | Green Ball | Blue Ball | Red matches |
| 3 | Blue Ball | Red Ball | Green Ball | Green matches |
| 4 | Blue Ball | Green Ball | Red Ball | None match |
| 5 | Green Ball | Red Ball | Blue Ball | None match |
| 6 | Green Ball | Blue Ball | Red Ball | Blue matches |
Counting Ways with No Matching Colours
We are looking for the number of ways where no ball goes into the box of its own colour. From the table above, we can see that only arrangements #4 and #5 satisfy this condition:
- Arrangement #4: Blue Ball in Red Box, Green Ball in Blue Box, Red Ball in Green Box. None match.
- Arrangement #5: Green Ball in Red Box, Red Ball in Blue Box, Blue Ball in Green Box. None match.
There are 2 ways to achieve this condition.
Understanding Derangements
This type of problem is known as finding the number of derangements. A derangement of a permutation is a permutation of the elements of a set, such that no element appears in its original position. The number of derangements of a set of \(n\) elements is often denoted by \(D_n\) or \(!n\).
The formula for the number of derangements of \(n\) items is:
\(D_n = n! \sum_{i=0}^n \frac{(-1)^i}{i!}\)
For \(n=3\) (three balls and three boxes):
\(D_3 = 3! \left( \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} \right)\)
\(D_3 = 3! \left( \frac{1}{1} + \frac{-1}{1} + \frac{1}{2} + \frac{-1}{6} \right)\)
\(D_3 = 6 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} \right)\)
\(D_3 = 6 \left( \frac{1}{2} - \frac{1}{6} \right)\)
\(D_3 = 6 \left( \frac{3}{6} - \frac{1}{6} \right)\)
\(D_3 = 6 \left( \frac{2}{6} \right)\)
\(D_3 = 2\)
Using the derangement formula confirms that there are 2 ways to place the balls such that no ball goes into the box of its own colour.
Conclusion
By listing the possibilities or using the formula for derangements, we find that there are 2 ways to put the three coloured balls into the three coloured boxes such that no ball is in the box of its matching colour.
Important Questions from Permutations and Combinations
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