For one survey 39 students from first, second and third year must be chosen, so that at least 8 students are from first year, and at least 3 students from second year. How many possible ways are there to make this? I tried to use the combinations formula, but I don't know how many students come from the first year, nor how many from the second. So I'm stuck.
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2What have you tried to solve the problem yourself? – Tom Finet May 21 '22 at 21:01
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I tried to use combinations, but it confuses me because there is not defined how many students are there from each year. – Pol May 21 '22 at 21:06
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1See Stars and Bars 1 and see Stars and Bars 2. Then, consider $$x_1 + x_2 + x_3 = (39 - 8 - 3).$$ – user2661923 May 21 '22 at 21:17
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Presumably the students from each year are indistinguishable – Henry May 21 '22 at 22:02
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The colleagues above introduces the concepts of combinatorics well. I will repeat the statement requirements. The notation for number of students by class is $n_0$, $n_1$ and $n_2$ for first, second and third year respectively.
The total number of students $n$ is 39. The minimum number of first year students $n_0$ is 8. The minimum number of second year students $n_1$ is 3. It means that the maximum number of chosen third year students is equal to 28.
Firstly, we need to find the number of partitions to number 28. The wikipedia article refers to 3718 manners to sum 28. We are interested on those composed by 3 summands, 0 inclusive. It means, it may look like $(0, 0, x_3)$, $(0, x_2, x_3)$ or $(x_1, x_2, x_3)$. You may run this utilitary function to obtain the total partitions and filter sufficient values out.
The total number of combinations is, for each tuple $(x_1, x_2, x_3)$, equal to the sum $C_{n_0, x_1+8} \, C_{n_1, x_2+3} \, C_{n_2, x_3}$, for notation $C_{i, j} = \frac{j!}{i! \, (i-j)!}$ corresponds to Newton binomial.
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