How many ways can I pick a size $3$ set from $\{1, 2, 3, 4, 5, 6, 7, 7, 7\}$ such that at least one is odd?

Question

I am trying to solve the following puzzle:

I have nine cards, where six of them are labelled $1$ through $6$ 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 $\{1, 2, 7\}$ count the same as $\{2, 1, 7\}$? I am assuming the order does not matter.

I believe there are $42$ possible card draws. I calculated this by considering three cases:

1. In the first case we draw three distinct integers and there are ${}_7C_3 = 35$ ways to do this.
2. In the second case we draw two $7$s and one non seven, there are $6$ ways to do this ($\{7, 7, 1\}, \{7, 7, 2\},\dots, \{7, 7, 6\}$).
3. In the third case we draw three 7s and there is only $1$ way to do this.

Summing the three cases gives $42 = 35 + 6 + 1$.

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 $42 - 1 = 41$.

Is my answer correct? Any thoughts or feedback is appreciated.

Answer 1

I have tried like this,

Case I: One 7 and two cards from {1,2,3,4,5,6}

$6C2=15$ ways.

Case II: Two 7's and two cards from {1,2,3,4,5,6}

$6C1=6$ ways.

Case III: Three 7's

$1$ way.

Case IV: Three cards from {1,2,3,4,5,6}

$3$ odds, $3$ evens. To get atleast 1 odd,

$3C3\times3C0[three\ odds]+ 3C1 \times 3C2[one\ odd] +3C2 \times 3C1[two\ odds]=19$

Hence, total possible ways is $15+6+1+19=41$ ways.