Combinations, help please

Question

$13$ 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 $1$ of these players must be a woman?

I think the answer is $3 \times \dbinom{12}{10}$. Am I right?

Answer 1

Hint: A standard approach is to (i) Count the number of choices, if there are no restrictions and (ii) Count the number of all-male teams.

Remark: Your current suggested answer is not correct. You have $3\binom{12}{10}$. I do not understand where the $10$ comes from, common (but wrong) reasoning would have a $9$. The reason that $3\binom{12}{9}$ is wrong is that it counts more than once teams that have more than one woman on them.

A correct version of your approach goes as follows. We can have $1$ woman, or $2$, or $3$. The number of $1$-woman teams is $\binom{3}{1}\binom{10}{9}$.
The number of $2$-woman teams is $\binom{3}{2}\binom{10}{8}$. And the number of $3$-woman teams is $\binom{3}{3}\binom{10}{7}$. Add up. More work than the other approach, specially since there are very few all-male teams.

Answer 2

The total number of ways to choose $10$ people from a pool of $13$ is $\dbinom{13}{10}$. The number of ways of choosing a team with at-least one woman is nothing but the number of ways of choosing a team with all men and subtracting it from the total number of ways. The number of ways of choosing a team with all men is $\dbinom{10}{10} = 1$. Hence, the number of ways of choosing a team with at-least one woman is $$\dbinom{13}{10} - 1 = \dfrac{13 \times 12 \times 11}{3 \times 2 \times 1} - 1 = 13 \times 2 \times 11 - 1 = 285$$

Answer 3

Count the complement: $\binom{13}{10} - \binom{10}{10}$