Blue and red balls
Probability
Level
3
You have 4 blue balls and 4 red balls.
How many ways can you label these balls with the numbers 1-8 such that no three similarly colored balls have two numbers that add up to the third number?
0
1
2
3
4
5
6
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1 solution
Since 1 + 2 = 3, balls 1-3 cannot be the same color, so there are two of one color and one of the other. For now we will consider two blues and one red, and later on we will double our findings to include the symmetrical case of two reds and one blue. For now this gives us 3 possibilities to consider: (1) RBB, (2) BRB, and (3) BBR.
Case (1) RBB : Since balls 2 and 3 are blue, ball 5 must be red. Since balls 1 and 5 are red, balls 4 and 6 must be blue. However, now balls 2, 4, and 6 are all blue, so this case is not possible.
Case (2) BRB : Since balls 1 and 3 are blue, ball 4 must be red. Since balls 2 and 4 are red, ball 6 must be blue. Since balls 1 and 6 are blue, balls 5 and 7 must be red. However, now balls 2, 5, and 7 are all red, so this case is not possible.
Case (3a) BBRR : Since balls 3 and 4 are red, ball 7 must be blue. Since balls 2 and 7 are blue, ball 5 must be red. Since balls 1 and 7 are blue, ball 8 must be red. However, now balls 3, 5, and 8 are all red, so this case is not possible.
Case (3b) BBRB : Since balls 1 and 4 are blue, ball 5 must be red. Since balls 2 and 4 are blue, ball 6 must be red. Since balls 3 and 5 are red, ball 8 must be blue. Since balls 1 and 8 are blue, ball 7 must be red. There are no contradictions with this sequence (BBRBRRRB) so this is a valid way of labeling the balls.
Starting with two blues and one red for balls 1-3 there is only one valid way of labeling the balls (BBRBRRRB). By symmetry, starting with two reds and one blue for balls 1-3 would give us one other valid way of labeling the balls (RRBRBBBR). There are therefore 2 ways of labeling the balls with the given conditions.