Let us begin by assigning 3 marks to each of the 10 questions, so that each will
have at least 3 marks.
We have now assigned 30 of the 50 marks. We have 20 more marks left to assign
to questions.
Get 20+10 or 30 slips of paper. (You don't have to actually get 30 slips of
paper. You can just imagine them, but I'll talk as though you actually have
30 slips of papers laid out in a row left to right).
Write the word "Divider #10" on the right-most slip of paper.
Now choose 9 other slips of paper from the other 29. You can choose them
C(29,9) ways.
Write "Divider #1" on the left-most one you chose. After Divider #1, going left
to right, write "Divider #2" on the next one chosen on the right, and so on
until the last slip chosen as a divider before Divider #10 will have "Divider
#9" written on it.
So now we have a row of 30 slips of paper. 10 of the slips of paper, including
the very last one, (the rightmost one), have "Divider #(something)" written on
them. And 20 slips of paper have nothing written on them.
Now going left to right, count the number of slips of paper that have nothing
written on them which are left of Divider #1, which may be anything from none at
all (zero) through 20. Whatever that number is, assign that number of extra
marks to question #1, (in addition to the 3 that have already been assigned to
it.
Next count the slips of paper between dividers #1 and #2, and assign that number
of marks to question 2.
Continue doing this until all 20 marks have been assigned.
[Remember that if dividers #n and #n+1 are next to each other, then you don't
assign any additional marks to question n+1.]
The number of ways we can pick the 9 slips of paper to be dividers (besides
divider #10 on the far right end) is C(29,9) = 10015005.
Answer: 10015005
Edwin
