I am trying to solve the following puzzle:
I have nine cards, where six of them are labelled 1$1$ through 6$6$ and the remaining three are indistinguishable and labelled with 7. Calculate the number of ways I can pick a set of three cards such that at least one is odd.
First of all, it's not clear to me if the order matters, i.e. does drawing {1, 2, 7}$\{1, 2, 7\}$ count the same as {2, 1, 7}$\{2, 1, 7\}$? I am assuming the order does not matter.
I believe there are 42$42$ possible card draws. I calculated this by considering three cases:
- In the first case we draw three distinct integers and there are 7C3 = 35${}_7C_3 = 35$ ways to do this.
- In the second case we draw two 7s$7$s and one non seven, there are 6$6$ ways to do this ({7, 7, 1}, {7, 7, 2}, ..., {7, 7, 6}$\{7, 7, 1\}, \{7, 7, 2\},\dots, \{7, 7, 6\}$).
- In the third case we draw three 7s and there is only 1$1$ way to do this.
Summing the three cases gives 42 = 35 + 6 + 1$42 = 35 + 6 + 1$.
Finally, there is only one way to have all even cards. Therefore, the number of ways I can pick a set of three cards such that at least one is odd is 42 - 1 = 41$42 - 1 = 41$.
Is my answer correct? Any thoughts or feedback is appreciated.
