There are ten contestants in an archery competition. Four of the contestants are women. Prizes are awarded to the top four competitors. If at least one woman finishes in the first four places, in how many ways can the top four places be filled?
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Tried figuring this out for about 45 minutes now, definitely gave it a crack, all help is very much welcomed and appreciated. – Josh Jul 31 '16 at 11:43
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This is too vague. Does the order matter? That is, is $WWWM$ different from $MWWW$? Are the men and women distinguishable? That is, is $W_1W_2xy$ different from $W_2W_1xy$? In any case, the numbers are so small, surely straight enumeration gets the job done, no? – lulu Jul 31 '16 at 11:47
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@lulu Did you just suggest brute force? – Kenny Lau Jul 31 '16 at 11:47
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@KennyLau Why, yes. Am i missing something? – lulu Jul 31 '16 at 11:48
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@lulu That's so against the spirit of mathematics. – Kenny Lau Jul 31 '16 at 11:48
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@KennyLau Ah, But for straight combinatorics with very small numbers, nothing beats enumeration. And then you have a solid example which you can compare against abstract methods. Nope, I'm sticking with enumeration. – lulu Jul 31 '16 at 11:50
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@KennyLau I got 10P4, which yields 5,040 ways (ignoring the terms of the question). – Josh Jul 31 '16 at 11:52
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@Josh You said order doesn't matter, which makes it 10C4 instead of 10P4 – Kenny Lau Jul 31 '16 at 11:55
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@KennyLau Order does matter, apologies. – Josh Jul 31 '16 at 11:57
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@KennyLau Question is really vague I know, sorry. – Josh Jul 31 '16 at 11:59
1 Answers
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If we ignore the condition that at least one woman finishes, we would come up with $P_4^{10}=5040$ possible ways.
Now, the opposite of the condition that at least one woman finishes, is that no woman finishes, which means that the 4 spots are all not women, which gives us $P_4^6=360$, with the $6$ because there are $4$ women, hence $6$ non-women.
Therefore, the answer is $P_4^{10}-P_4^6=4680$.
From the simple Venn diagram below, one can see that:
$$\mbox{Regardless of the condition}=\mbox{Opposite of the condition}+\mbox{According to the condition}$$
Kenny Lau
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