For the 6 faces and 8 vertices of a regular hexahedron, respectively sealed with red and blue these 2 beads (each face contains a bead and each vertex contains a bead), how many different types of solutions are there?
To determine the number of different types of solutions for the colored beads on the regular hexahedron, we can break the problem down into two parts:
1. Counting the number of ways to arrange the red beads on the faces of the hexahedron.
2. Counting the number of ways to arrange the blue beads on the vertices of the hexahedron.
Let's look at each part separately:
1. Arranging the red beads on the faces:
- Since there are 6 faces on the hexahedron and each face can have one red bead, we need to count the number of ways to distribute 2 red beads among 6 faces.
- To solve this, we can use a combination formula. The number of ways to distribute 2 indistinguishable objects (the red beads) among 6 distinct boxes (the faces) is given by the combination formula: C(6, 2) = 6! / (2! * (6-2)!) = 15.
2. Arranging the blue beads on the vertices:
- Since there are 8 vertices on the hexahedron and each vertex can have one blue bead, we need to count the number of ways to distribute 2 blue beads among 8 vertices.
- Again, we can use the combination formula. The number of ways to distribute 2 indistinguishable objects (the blue beads) among 8 distinct boxes (the vertices) is given by the combination formula: C(8, 2) = 8! / (2! * (8-2)!) = 28.
To find the total number of different types of solutions, we multiply the number of solutions for each part together: 15 * 28 = 420.
Therefore, there are 420 different types of solutions for the colored beads on the regular hexahedron.