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I've tried to solve the following exercise (Ross First Course 8th edition):

I have to chose three students from three groups of $n$ students (in total exist $3n$ students):

If I must choose two students of the same group and the other from a different class? My try: First I chose a class of the three, then two students of that class, a class from two other class and finally an student: $$3 \binom{n}{2} 2 \binom{n}{1}$$.

Then the next exercise is: If emphasized textthe three students must be in different classes My try: $3 \binom{n}{1} 2 \binom{n}{1} 1 \binom{n}{1}$. But the solution in the textbook is $n^3 = {\binom{n}{1}}^3$.

What's the difference?

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    $\begingroup$ Because you're choosing 3 groups you have $3\choose 3$$=1$ ways to choose 3 groups and n ways to select a student from each $\endgroup$ – user10444 Jan 20 '14 at 20:50
  • $\begingroup$ But why in the first exercise I have to take in account the order? Or why first I can choose three and the two? $\endgroup$ – David Jan 20 '14 at 21:21
  • $\begingroup$ Because you can choose group ($1$ and $2$) or ($2$ and $3$) or ($1$ and $3$). While in the last one you can choose all groups in only one way. $\endgroup$ – user10444 Jan 20 '14 at 21:24
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    $\begingroup$ To emphasize more when you write $3$ what is meant is $3 \choose 1$ because you are finding in how many ways can you get 1 group from 3. Now you are left with 2 groups and you want one of them that can be done in $2 \choose 1$ ways. $\endgroup$ – user10444 Jan 20 '14 at 21:28
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Compare these choices:

  • Alice from class A,
  • Bob from class B,
  • Charlie from class C,

with

  • Bob from class B,
  • Charlie from class C,
  • Alice from class A.

You've counted them as different choices, whereas the text considers them the same (the factor of $3 \times 2 \times 1$ represents the order in which the classes are chosen).

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  • $\begingroup$ Sorry, but can explainme why in the first exercise if I choose two from a class $\binom{3 $\endgroup$ – David Jan 21 '14 at 6:10
  • $\begingroup$ It's slightly different in the earlier case: Compare "Alice and Andrew from class A and Bob from class B" vs. "Alice from class A and Bob and Brenda from class B". These are different. I.e., choosing two things from class A and one from class B is different to choosing two things from class B and on from class A. $\endgroup$ – Rebecca J. Stones Jan 21 '14 at 6:15
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The problem with your solution is that $3n\cdot2n\cdot 1n$ implies that you care in which order you select the students: there is a first, a second, and a third.

So, this overcounts: choosing students A, B, and C is represented by the orders of choosing ABC, ACB, BAC, BCA, CAB, CBA.

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Because you're choosing $3$ groups you have $3\choose 3$$=1$ way to choose $3$ groups and $n$ ways to select a student from each. You get $3 \choose 3 $$n \times n \times n=n^3$. We don't multiply by $3$ and $2$ because we are not interested from what specific group that student comes from.

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