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?


The answer provided is simply wrong. The correct answer is


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


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

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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