I feel that I have the correct expression for this problem, but I have this nagging feeling that something is incorrect. May I ask that you all verify this?
Question: "In a game of chance, three of the same color wins. Suppose there are thirty colored coins of the same size in a bag. The bag contains an equal amount of red, blue, green, yellow and white colored coins. If a player is allowed only one draw of three coins, how many possible winning draws are there? How many ways can the player lose?"
My Answer:
- C(30, 3)
- C(30, 27)
There are $6$ of every colored coin. To win, you can choose exactly $3$ red coins. There are $\text C(6, 3)$ ways to do that. But you can win with any of the other $4$ colors, so there are $5\cdot \text C(6, 3)$ ways to win. Any other draw you make will result in a loss, so the remaining $\text C(30, 3)-5\cdot\text C(6, 3)$ draws are ways to lose.