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  1. \({7 \choose 2}\) solutions. After each variable gets 1 star for free, we are left with 5 stars and 2 bars.

  2. \({10 \choose 2}\) solutions. We have 8 stars and 2 bars.

  3. \({19 \choose 2}\) solutions. This problem is equivalent to finding the number of solutions to \(x' + y' + z' = 17\) where \(x'\text{,}\) \(y'\) and \(z'\) are non-negative. (In fact, we really just do a substitution. Let \(x = x'- 3\text{,}\) \(y = y' - 3\) and \(z = z' - 3\)).

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