$13$ 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?

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You have asked $7$ questions so far but have not accepted answers to any of them. –  user17762 Nov 3 '12 at 20:55
Hey, I am active user in stackoverflow and I answer many programming questions I expected some one would help me in math as I help many in programming ! –  Alpha Nov 3 '12 at 21:07
If the answers that you get aren’t useful enough to warrant being accepted, why do you keep coming back? I’m more relaxed about this than a lot of the regulars, but even I’m starting to get a little irked. And this profile doesn’t support your claim about answering many questions in StackOverflow. –  Brian M. Scott Nov 3 '12 at 21:15
Huh.... Didn't know that. (Certainly not something I'd've expected to learn on a math site!) –  Cameron Buie Nov 3 '12 at 21:45
I am downvoting your question because you have posted several homework questions without displaying any effort of your own. Moreover, you fail to adhere to the community policy of accepting answers. Please show us you are willing to be a valuable member of this community by asking thoughtful questions and supporting those that take the time to assist you. –  Austin Mohr Nov 3 '12 at 21:49

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.

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so whose answer is correct ?!!! I am Confused.. complementing correct or.. –  Alpha Nov 3 '12 at 21:12
@Alpha: The answers given are both right. My hint would lead to the same thing. The answers were downvoted (in my opinion wrongly) because they gave full answers. I will upvote them, but unfortunately that will not fully correct the downvotes. –  André Nicolas Nov 3 '12 at 21:13
Thanks man,appreciated –  Alpha Nov 3 '12 at 21:14
@Alpha If you found this a correct and useful answer, you should probably accept it. –  Neal Nov 5 '12 at 16:39
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 $$\dbinom{13}{10} - 1 = \dfrac{13 \times 12 \times 11}{3 \times 2 \times 1} - 1 = 13 \times 2 \times 11 - 1 = 285$$
Count the complement: $\binom{13}{10} - \binom{10}{10}$