Question

some friends are playing a game with two spinners. each spinner is spun once. to win the game, you must spin a different color on each spinner. how many ways are there to win the game? \square ways

Explanation:

Step1: Analyze spinner colors

Each spinner has 4 colors: blue (B), green (G), red (R), yellow (Y).

Step2: Count winning combinations

For first spinner color B: second spinner can be G, R, Y (3 ways).
For first spinner color G: second spinner can be B, R, Y (3 ways).
For first spinner color R: second spinner can be B, G, Y (3 ways).
For first spinner color Y: second spinner can be B, G, R (3 ways).
Total ways: \( 4 \times 3 = 12 \)? Wait, no—wait, each spinner is identical? Wait, no, wait: each spinner has 4 sections. Wait, no, let's list all pairs. First spinner (S1) and second (S2). S1 has 4 colors, S2 has 4. Total possible: \( 4 \times 4 = 16 \). Losing is same color: (B,B), (G,G), (R,R), (Y,Y) → 4 losing. So winning: \( 16 - 4 = 12 \)? Wait, no, wait the spinners are identical? Wait the first spinner: blue, green, red, yellow. Second same. So for each color in S1, S2 must be different. So for S1=B: S2 can be G, R, Y (3). S1=G: S2=B, R, Y (3). S1=R: S2=B, G, Y (3). S1=Y: S2=B, G, R (3). Total: \( 3 + 3 + 3 + 3 = 12 \). Wait, but maybe I made a mistake. Wait the spinners are divided into 4 equal parts, same colors. So each spinner has 4 options. So the number of ways to get different colors is total possible (4×4=16) minus same color (4) → 12. Or directly, for each of the 4 colors in first spinner, 3 in second. So 4×3=12.

Answer:

12