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I have been struggling with this question for over a day. I think I am on the correct track, but I am not certain.

My approach was to use stars and bars for the representation of the team. I places the first bar so that the total to the left was exactly 2 and counted the spaces that the second bar could go while leaving the right most segment with at least 2 players.

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The total ended up being 5, so I have P(5,1) places for the bar to go. I repeated this process moving the initial bar one space to the right each time and summed the results, which I found to be P(5,1)+P(4,1)+P(3,1)+P(2,1)=5+4+3+2=14.

Am I correct or have I made a mistake? Thank you for your help. This is my first semester of combinatorics and the first time I have done math like this.

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$14$ is the wrong answer. We have two players reserved for each of the defender, midfielder and forward rows, so we might as well ignore them. We have $4$ players left to distribute however we like among the three positions – this is where we apply stars and bars: 4 stars, 2 bars, $\binom{4+2}2=\binom62=15$.

Note that some formations like 4-2-3-1 are completely unrepresented by this format.

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  • $\begingroup$ I'm so sorry, your answers is great, but I am afraid I am missing one minor detail to fully understand it. I am a little slow on why we are representing the distribution of the final players as P(4+2,2). $\endgroup$
    – Goldsten
    Jan 9, 2020 at 16:55
  • $\begingroup$ @Goldsten More properly, the calculation is $\binom{4+3-1}4$, there being $4$ leftover players and $3$ positions. $\endgroup$ Jan 9, 2020 at 16:57

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