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Twenty people are called for jury service. Four of them are students, seven are waged employees, five are self-employed and four are retired. (a) How many different twelve-person juries can be selected? (b) How many different compositions of juries can be selected? (Two juries have the same composition if they have the same numbers of students, waged employees, self-employed and retired people.)

Working: (a) 20C12 = 125970 (I think)

(b) I worked out the number of juries containing 3 of each as 4C3 x 7C3 x 5C3 x 4C3 = 5600. 125970 - 5600 = 120370 (seems too big to be right)

N. F. Taussig
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    Find no. Of nonnegative solutions to $x_1+x_2+x_3+x_4=12$ with some restrictions on the $x_i$s. – insipidintegrator Jul 02 '22 at 15:51
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    The first part is good, I can't understand what you wrote for the second part. Note that "$3$ of each type" is one composition. No need to count how many juries have that particular composition. – lulu Jul 02 '22 at 15:52
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    Your second answer addresses the wrong question. As @lulu points out, a composition can (will) have many, many possible combinations of jurors. You're counting combinations when you're supposed to be counting compositions. The approach suggested by insipidintegrator will work, but requires a bit of care to implement properly. – Robert Shore Jul 02 '22 at 17:49
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    For the generic problem, which Robert Shore has answered, for your particular posting, you need to combine the use of Inclusion-Exclusion, with Stars and Bars. See this answer. – user2661923 Jul 02 '22 at 18:58

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Let $x_1, x_2, x_3, x_4$ be the number of students, waged employees, self-employed, and retired members of your jury, respectively. Each different solution to $x_1+x_2+x_3+x_4=12$ in non-negative integers, subject to the constraints that $x_1 \leq 4, x_2 \leq 7, x_3 \leq 5, x_4 \leq 4$, is a different acceptable composition. How many such solutions are there?

First let's ask an easier question. How many solutions are there if we ignore the constraints? The starts-and-bars technique tell us that there are $\binom {15}{3}=455$ unconstrained solutions to this equation. But this is too big to be the correct answer because, for example, we're including combinations where all $12$ jurors are students, and there aren't that many available students.

So we have to subtract off the potential compositions that violate one or more of the constraints. We will do this using the principle of inclusion and exclusion.

How many potential solutions are unacceptable because they violate the first constraint, $x_1 \leq 4$? Let's find the number of solutions to $y_1+x_2+x_3+x_4=7$, where $y_1=x_1-5$. We can again use stars-and-bars to see that $\binom{10}{3}=120$ of our potential solutions are unacceptable because they violate the first constraint.

Similarly, $\binom 73=35$ potential solutions violate the second constraint, $\binom 93=84$ potential solutions violate the third constraint, and $\binom{10}{3}=120$ potential solutions violate the fourth constraint. Add these up and it seems that we should subtract off $359$ unacceptable compositions.

But that's not right because we've overcorrected. We have subtracted off the potential compositions with $5$ students and $5$ retirees twice, once because they violate the first constraint, and a second time because they violate the fourth constraint. That's where the principle of inclusion and exclusion comes in.

We need to determine how many potential combinations violate at least two constraints. Using similar reasoning to that stated above, we see that there are $4$ compositions that violate the first and third constraints, $4$ compositions that violate the third and fourth constraints, and $10$ compositions that violate the first and fourth constraints, for a total of $18$ compositions that violate at least two constraints.

We're done, because it's not possible to create a jury that violates three constraints. Thus, the correct answer is $455-359+18=114$ acceptable jury compositions.

Robert Shore
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