# Box Labels

Here's a puzzle.

Each of the three boxes contains 2 balls. In one there are two white balls, in another two black balls, and in the third one ball is black, the other is white. Boxes have been labeled to indicate their contents. However, whoever did the job got all labels wrong. The task is to straighten things out. You may select 1 box and blindly pick up a ball out of it.

In the applet, select box. One of the two balls it contains will pop up. After which you'll have to drag labels (they will appear at the right moment) into the empty label locations on the boxes.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Solution

## Solution

Click on the box (incorrectly) labelled white and black. It is bound to contain two balls of the same color, for otherwise the original label would be right. Verify that once the right label for that box is found, there's only one way to place the remaining labels.

Selecting a box labelled by two balls of the same color may or may not give you a clue to the right contents. For example, assume you pick the box labelled black/black. If you are lucky and out pops a black ball, you may be sure of the content: the box contains balls of different colors. But if, per chance, the shown ball is white, you get no useful information about what is inside.

That puzzle is very well known and probably is very old. A novelty could be in increasing the number of balls, boxes, and guesses. With three balls, one can label 4 boxes. (Using only the first letters, the labels are WWW, WWB, WBB, BBB, which naturally corresponds to 4 boxes.) Unfortunately, the puzzle becomes cumbersome as neither 2 nor 3 guesses may suffice to determine the correct labeling. 2 guesses never suffice. 3 guesses and some luck (i.e., given suitable popouts) may lead to a solution in some cases. What about 4 guesses?

(Permutations in which none of the elements remains in its place are called derangements. With 3 boxes, there were only 2 derangements. With 4 boxes, there are 9. Derangements are relevant to the puzzle since when we try to determine the contents of the boxes, we actually seek a derangement of a given (wrong) labelling.)

[Sharygin, p. 82] places a problem in a fanciful context:

Three pairs were dancing in a dance class. It so happened that the three girls had dresses of different colors: red, green and blue. The same was true of the three boys. The boy in the green suit was dancing with the girl in the red dress. What color dress was on the partner of the boy in the red suit?

### References

1. D. Rohrer, More Thought Provokers, Key Curriculum Press, 1994, #19
2. I. F. Sharygin, Mathematical Mosaic, Mir, 2002 (in Russian)