InclusionExclusion and Generating Function with Coefficient (and Integer Equation)
 The password contains at least one number AND
 At least two of the five special characters are consecutive
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Problem 2: Suppose we want to assign 46 computer science students to work on teams to complete 12 projects. Note that it is possible for some projects to receive no qualified students to work on the project.
We have the following restrictions on the number of students on the teams:
 Project teams 3 and 7 have exactly 8 students.
 Project teams 1, 2, and 12 have no more than 5 students.
 Project teams 4, 9, 10, and 11 have at least 3 students, but strictly less than 9 students.
 Project teams 5, 6, and 8 have at least 4 students but strictly less than 16 students.
 Modeling this combinatorial problem as an integer equation with restrictions
 Constructing a generating function
 Computing the relevant generating function coefficient
Answer
 The questioner was satisfied and accepted the answer, or
 The answer was disputed, but the judge evaluated it as 100% correct.

Sorry, in problem 1 I obviously meant that S2 and S4 are the sets with no number, not the ones with at least a number, plus the remaining conditions.

Similarly, in S3 and S4 I meant that no two special characters are consecutive, not that no two special characters are consecutive. These are the sets that should be removed.

*not that of which at least two are consecutive.

Summarising: S1 counts the strings that have 5 special characters, one in first position; S2 counts the strings that have 5 special characters, one in first position, and no number; S3 counts the strings that have 5 special characters, one in first position, and of which no two are consecutive; S4 counts the strings that have 5 special characters, one in first position, of which no two are consecutive, and no number.

Hi, thank you for your answers! Would you mind checking your answer to Problem 1 again? I'm computing both the "number of choices" solution and the inclusionexclusion sets, and I'm getting two different answers. Also, where did the 9!/5! and \binom{49}{4} come from?

You're right, the 9!/5! should be 9!/4!, I copied it incorrectly. That comes from the choices of the 5 distinct special characters, in order in which they appear in the password (9 choices for the 1st, 8 for the 2nd, etc). The (19 choose 4) comes from the choices of the positions of the special characters: one is fixed (the 1st one should be in 1st position), so there are 19 positions left for the remaining 4 characters. Is it clear?

Ah, I actually wrote a (49 choose 4), missed it. It should obviously be (14 choose 4), that is a typo.

I'm sorry for the typos, but when one has to answer quicky, they inevitably happen, even if one doublechecks everything. :/
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