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So a soccer team is creating a team of 11, out of 22 players. There are 4 positions, which a certain number of people want: 2 wants to play keeper, 8 wants to play defense, 8 wants to play middle field and 4 wants to play offense. The number of spots available per position is: 1 for keeper, 4 for defense, 4 for middle field and 2 for offense.

The conditions are these: Mary and Sue only wants to join the team if both get the middle field position. Francis only wants to join the team if the Harold doesn't get the middle field position. Lastly one of the candidates for the offense position, Chris, rather wants to join the middle field position.

So there are a total of 9 players who wants the middle field position.

The question is: Assuming that the coach picks a team according to everyone's preferences, will Chris have a better chance of joining the team if he switches his preferences to the middle field, from offense?

I've tried a number of ways to attack this problem, I figured the amount of combinations for one player to join the middle field is 56 out of 126 possible. I'm struggling with how to find the amount of combinations for Mary and Sue.

I don't necessarily need the total answer, just a hint or a reference to a rule would be of much help.

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  • $\begingroup$ I think we (and Chris) need to know more about just how the coach will "pick a team according to everyone's preference" in order to decide whether he should switch preferences. $\endgroup$ – Ethan Bolker Feb 3 '18 at 14:17
  • $\begingroup$ Are we to assume that the coach accepts the conditions some of the players are putting ? $\endgroup$ – true blue anil Feb 3 '18 at 14:23
  • $\begingroup$ The coach picking according to preferences means that he picks: 1 keeper out of 2, 4 defense out of 8, 4 middle field out 9, 2 offense out of 3. In addition, both Mary and Sue are getting middle field positions, and Francis gets a middle field position while Harold doesn't get a middle field position. $\endgroup$ – Doe J Feb 3 '18 at 14:27
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Using initials for named persons, the mid field has to be composed from [MS], [F or H], [4 (+ C)]

If C is to be selected for midfield, the $3$ others can be chosen in these ways:

  • $3$ from the last group:$\binom43 = 4$
  • $2$ from the last group and $1$ from the second group: $\binom42\binom21 = 12$
  • $1$ from the last group and $2$ from the first group:$\binom41\binom22 = 4$
  • Total ways of forming midfield if C is included = $4+12+4 = 20$

If C is not selcted for midfield, four others have to be chosen in similar fashion,
total ways of forming midfield = $\binom44 + \binom43\binom21 + \binom42\binom22 = 15$

Can you now proceed to a conclusion ?

Btw, you have used "odds" at one place, "chance" at another.
Pl. don't use "odds" as a synonym for "probability", they are not the same.

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  • $\begingroup$ Ah, in my haze of confusion I must've gotten "odds" and "chance" mixed up, thank you for the correction. As for the task: I tried to do the total chance for each of the restrictions: $One random player$: 9C4 - 8C4 = 56. $Mary and Sue$: (9C4 - 8C4) - (8C3 - 7C3) = 56 - 35 = 21. Etc. And then add together those answers to get the total number of combinations. I now see where I went wrong: I should've taken the combinations for each group in question, rather the entire selection. And by adding combinations + C (favorable outcomes) and dividing them by combinations - C, you get the $chance$ for C. $\endgroup$ – Doe J Feb 3 '18 at 15:48
  • $\begingroup$ Thank you so much for your precise, and well formulated answer! $\endgroup$ – Doe J Feb 3 '18 at 15:49
  • $\begingroup$ You are welcome ! $\endgroup$ – true blue anil Feb 3 '18 at 15:51
  • $\begingroup$ Odds in favor of C being selected for midfield = 20:15, Pr(C selected for midfield) = 20/35 $\endgroup$ – true blue anil Feb 3 '18 at 15:59

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