Preselection tests for the full-time training

Question

Ten students take a test consisting of 4 different papers in Algebra, Geometry, Number Theory and Combinatorics. First, the proctor distributes randomly the Algebra paper to each student. Then the remaining papers are distributed one at a time in the following order: Geometry, Number Theory, Combinatorics in such a way that no student receives a paper before he finishes the previous one. In how many ways can the proctor distribute the test papers given that a student may for example finish the Number Theory paper before another student receives the Geometry paper, and that he receives the Combinatorics paper after that the same other student receives the Combinatorics papers.

Accepted Answer

First, since the proctor distributes randomly the Algebra paper to each student, he has

10!

ways to do it depending on how he orders the students.

For the other three papers, we order the students from

1

to

30

, each student receiving three positions corresponding to the three papers. Since the first position a student receives corresponds to the Geometry paper, the second to the Number Theory paper and the third to the Combinatorics paper, the problem is equivalent to count the number of partitions of the set

{1, 2, \dots, 30}

into

10

subsets, each of

3

elements. This is equal to

(3 , \dots , 3 30) = \frac{30 !}{3 ! ^{10}} .

Therefore, the number of ways the proctor can distribute the test papers is

\frac{10 ! 30 !}{3 ! ^{10}} .