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This was a question on a recent test and I was hoping for a conclusive answer and reasoning behind it.

A local university housing office has a problem. It has 11 students to squeeze into 3 dorm rooms. It has been decided that 3 students are to be assigned to the first room, 6 students are to be assigned to the second room and 2 students are to be assigned to the third room. In how many ways can this assignment of 11 students be accomplished?

[Edit: As I recall,] the answer provided was: $11 \choose 3$ + $8 \choose 6$ + $2 \choose 2$ = 194

What I don't understand is why order matters in choosing how many students are assigned to each dorm. That is, why should the answer be different if 6 students are chosen for the first room and 3 chosen for the second?

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The answer provided is simply wrong. The correct answer is

$$\binom{11}3\binom86\binom22=4620\;.$$

And you can verify that this does not change if you assign choose the six students first and then the three, or in any other order. It’s simply

$$\frac{11!}{3!6!2!}\;.$$

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    $\begingroup$ Related: Multinomial coefficients allow you to write the answer symmetrically as $\binom{11}{3,6,2}$. $\endgroup$ – TMM Apr 30 '13 at 19:30
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Look at the problem in a different form.

Line up the 11 homeless students in some arbitrary order: birth-date, weight, GPA, alphabetical, whatever.

But first, take 11 cards, six with the word BIG written on them, 3 with MEDIUM, and 2 with SMALL. Shuffle the cards. There are $11!$ ways to arrange the cards, but rearranging the 6 BIG, or the 3 MEDIUM, or the 2 SMALL does not change the shuffled deck: there are $$\frac{11!}{6!\times 3!\times2!}$$different decks possible. Then just hand out the cards to the students in their line.

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  • $\begingroup$ Very nice, can you provide some more examples like these? $\endgroup$ – CodeYogi May 4 '16 at 7:54

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