Distribute the Adjacent Circles Activity Sheet to each student. Students will be coloring the patterns to determine the total number of possibilities. Each student will need a red, blue, and yellow crayon.
Adjacent Circles Activity Sheet
Individually, students should read the scenario on the activity sheet. Before beginning the activity, they should make a prediction and record it on the activity sheet. It may be necessary to discuss the situation with students before they begin, to ensure that all students understand the problem. You might want to read the problem aloud to students:
Five circles are connected as shown. Each of the five circles must be colored solidly with one of three colors: red, blue, or yellow. Two circles connected by a line segment may not be the same color.
Then, draw a copy of the figure on the overhead projector or chalkboard and demonstrate one way that the figure could be colored. For instance, using the figure below as an example, explain that A could be colored blue, but then neither B nor C could be colored blue. Continue until all five circles have been discussed.
Allow enough time for students to color and to compare their prediction to the actual total.
In groups of two or three, students should compare their predictions and actual total number of possibilities. Circulate among the groups, and offer assistance as necessary. In particular, it is quite likely that students will have missed some possibilities, or they may count some patterns more than once. Help them to create a system for organizing their possibilities. For instance, the solution shown below organizes the possibilities into two broad categories: one where B and C are the same color, and the other where B and C are different colors; within the first category, the list is further organized into three groups, one for each of the repeated colors.
Students should determine that there are 36 different combinations possible. As shown by the organization above, the problem is best attacked by considering two cases: 1) where B and C are the same color, and 2) where B and C are different colors.
Although students at this grade level may not be sophisticated enough to understand the following "proof," it does provide a way to verify the answer combinatorially.
Case 1 — B and C are the same color. There are 3 choices of color for B and C. Then, both A and D can be either of the other two colors. Finally, E must be one of the two colors that is different from D. In total, then, there are 3 × 2 × 2 × 2 = 24 possibilities when B and C are the same color.
Case 2 — B and C are different colors. There are 3 choices for B and 2 choices for C. But once those colors are chosen, then the color of both A and D are determined — both must be the color not yet used. Finally, there are 2 choices for the color of E, since it must be different from the color of D. In total, that gives 3 × 2 × 1 × 1 × 2 = 12 possibilities when B and C are different colors.
Altogether, then, there are 24 + 12 = 36 possibilities.
Record the following chart on the chalkboard or overhead. Ask students to copy and complete the chart for the same configuration as on the activity sheet.
Number of Colors
Number of Possibilities
As shown above, there are 36 possibilities when 3 colors are used. For 1 color, there are 0 possibilities, and with 2 colors, there are 2 possibilities.
Marcy Cook. "IDEAS: Combinations." The Arithmetic Teacher. 36, 1 (September 1988) 31-36.
Ask students to determine a general rule that they can write algebraically. Students can work in their groups to determine their rule. Students may add the number 4 to the previous chart, in which case they should get 252 possibilities.
[Case 1 — B and C are the same color. There are 4 choices of color for B and C. Then, both A and D can be either of the other 3 colors. Finally, E must be one of the 3 colors that is different from D. In total, then, there are 4 × 3 × 3 × 3 = 108 possibilities when B and C are the same color.
Case 2 — B and C are different colors. There are 4 choices for B and 3 choices for C. Once those colors are chosen, then there are 2 choices for the color of either A and D — both must be a color not yet used. Finally, there are 3 choices for the color of E, since it must be different from the color of D. In total, that gives 4 × 3 × 2 × 2 × 3 = 144 possibilities when B and C are different colors.
Altogether, then, there are 108 + 144 = 252 possibilities when four colors are used.]