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I'm wanting to know if my working for this question seems reasonable, or if there is a better way to do it. The question is as follows. A final exam has 12 questions, the marks for which must sum to 200. Each question should be worth at least 10 marks but less than 25. Moreover, the marks for each question should be divisible by 5. In how many ways can the marks be assigned?

I first setup this equation:

x1 + x2 + .. + x12 = 200, 10 <= xi < 25

Which becomes:

x1 + x2 + .. + x12 = 80, 0 <= xi <= 14

Now, to get the solution to this equation I find the number of solutions where x is unbounded, then remove solutions where 1,2,3 .. x's exceed 14. Something like (sorry, I'm not familiar enough with LaTeX to use it yet):

C(91, 80) - C(12, 1) * C(76, 15) - ... C(12,5) * C(16, 5)

I then consider the limitation that the marks must be divisible by 5. In the set from {0, .. 14} there are only 3 numbers divisible by 5: 0, 5, 10. As such I then multiplied my result above by 3/15 to get the number of solutions.

Now, my question. Is this working even correct? If not where did I go wrong and what would be the correct way to do it?

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In $\LaTeX$ write x_{12} for $x_{12}$, \le for $\le$ and \binomial{a,b} or {a \choose \b} for $\binom{a}{b}$ –  Henry Apr 25 '12 at 0:07

1 Answer 1

up vote 0 down vote accepted

Your first step was sensible going from the number of solutions to $$x_1 + x_2 + .. + x_{12} = 200, 10 \le x_i \lt 25$$ to the number of solutions to $$y_1 + y_2 + .. + y_{12} = 80, 0 \le y_i \le 14.$$

You should take the multiple of $5$ now rather than later to get the number of solutions to $$z_1 + z_2 + .. + z_{12} = 16, 0 \le z_i \le 2$$ as otherwise you will get a number which is far too high.

Just looking at partitions of 16 into up to 12 parts each no more than 2, there are only five possibilities:

  • $2+2+2+2+2+2+2+2+0+0+0+0$
  • $2+2+2+2+2+2+2+1+1+0+0+0$
  • $2+2+2+2+2+2+1+1+1+1+0+0$
  • $2+2+2+2+2+1+1+1+1+1+1+0$
  • $2+2+2+2+1+1+1+1+1+1+1+1$

but these can come in any order so the number of possibilities is

$$\frac{12!}{8! \times 0! \times 4!} + \frac{12!}{7! \times 2! \times 3!} + \frac{12!}{6! \times 4! \times 2!} + \frac{12!}{5! \times 6! \times 1!} + \frac{12!}{4! \times 8! \times 0!} = 28314$$

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