In general, the sum of the numbers in the second column equals twice the number of lines, since each number in each of the two columns must be counted once. So for 4 lines, a + b + c + d = 8 (where a is the number of 1’s, b is the number of 2’s, etc.).

Also, we can multiply the number of 1’s in the second column (a-1, since there is one in the first column) by 1, add it to the number of 2’s in the second column (b-1) multiplied by 2, and so on. This must also equal 8.

1*(a-1)+2*(b-1)+3*(c-1)+4*(d-1)=8

a + 2b + 3c + 4d = 18

Since I only have two equations for four variables, I had to use some trial and error to get two solutions for four lines, which are:

(a,b,c,d) = (2,3,2,1) and (a,b,c,d) = (3,1,3,1)

For five lines, a+b+c+d+e=10 and a+2b+3c+4d+5e=25 and I found one solution:

(a,b,c,d,e) = (3,2,3,1,1)

I am interested to know if there are more precise general principles than my two equations that would not make it unpleasant to look for solutions beyond 6 lines. [/spoiler] ]]>

The number of 2?s in this quiz is 3

The number of 3?s in this quiz is 2

The number of 4?s in this quiz is 1 Quiz B

The number of 1?s in this quiz is 3

The number of 2?s in this quiz is 1

The number of 3?s in this quiz is 3

The number of 4?s in this quiz is 1

[/spoiler] ]]>

http://pastebin.com/6rhxv0yA

the last guy gets his seat about 0.499045 of the time.

]]>