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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.

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    $\begingroup$ You must multiply $C(3,1)$ with $C(12,9)$ because for every choice of a woman you have $C(12,9)$ choices for the other people. But this does not work either because you count arrangements multiple times. $\endgroup$ – Peter Jun 12 '16 at 22:15
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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 ] $\;=\dbinom{13}{10} - \dbinom{10}{10}$

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  • $\begingroup$ I get what you are saying, but the constraint is "at least one", so wouldn't having a combination where Alicia is picked first, followed by a combination where Britney is picked and then Alicia is picked, be perfectly valid? Because there is still "at least one" in each combination. Unless I misunderstood you. $\endgroup$ – Dave Jun 12 '16 at 22:51
  • $\begingroup$ $A$ followed by $B$, and $B$ followed by $A$ are valid, but you are counting them twice (apart from other overcounts). Problems of the "at least one" type are best tackled using the complement. You can verify for yourself that the book method, and the above method give the same answer. $\endgroup$ – true blue anil Jun 13 '16 at 4:44
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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.


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 $$\binom31 \binom{10}9+\binom 32\binom {10}8+\binom33\binom {10}7=285$$

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    $\begingroup$ This is a better solution than the one on the supplied answer sheet. On the other hand, it doesn't explain why OP's method fails. $\endgroup$ – David K Jun 12 '16 at 22:18

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