How to pick $10$ people from $13$ such that at least $1$ is a woman

Question

Problem 10c from here.

Thirteen people on a softball team show up for a game. Of the $13$ people who show up, $3$ are women. How many ways are there to choose $10$ players to take the field if at least one of these players must be a woman?

The given answer is calculated by summing the combination of $1$ woman + $9$ men, $2$ women + $8$ men, and $3$ women + $7$ men.

My question is, why can't we set this up as the sum $\binom{3}{1} + \binom{12}{9}$ - picking one of the three women first, then picking $9$ from the remaining $12$ men and women combined? The only requirement is that we have at least one woman, which is satisfied by $\binom{3}{1}$, and that leaves a pool of $12$ from which to pick the remaining $9$. The answer this way is close to the answer given, but it's $62$ short. I get that it's the "wrong" answer but I'm wondering why my thinking was wrong in setting it up this way. Thanks.

Answer 1

Your answer is wrong on two counts:

Firstly you multiply, not add, if you are thinking of $\dbinom31$ and $\dbinom{12}{9}$

Secondly, this approach will over count. Suppose you chose Alicia , and then you chose $9$ from the remaining $12$, you would also have combos where Britney was first chosen, and Alicia was chosen from the $12$ group.

Thirdly, your book approach is correct, but unnecessarily tedious.

The best way is to compute
[All possible combos] - [All male combos ] $\;=\dbinom{13}{10} - \dbinom{10}{10}$

Answer 2

Among $13$ people, you can choose $10$ players in $$\binom {13}{10}$$ ways.

Now, suppose there is no woman in the squad, then you can choose that team in $\binom{10}{10}=1$ way.

So, number of ways of choosing a team with at least $1$ woman is $$\binom{13}3-1=286-1=285$$ ways.

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When you are counting several countings(call it nested counting), you should apply the multiplication principle. Here counting ways of choosing $1$ woman and $9$ men, you are in some kind of this nested counting, where, for any $9$ men among the $10$, the woman can occur, so, have to multiply this.

And upon deriving each choices by multiplying, you add those terms. Because, in one counting, there is $1$ woman , in some other count, there are $2$, and in other, there are $3$, women.

So, all are separate cases, so add them, like $$\binom31 \binom{10}9+\binom 32\binom {10}8+\binom33\binom {10}7=285$$