# How many ways an examiner can assign 15 marks to 5 questions?

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This is a straightforward stars-and-bars question. Since each question must receive at least two marks, you have $15-2\cdot5=5$ marks left to distribute as you please amongst the $5$ questions, so you’re putting $5$ indistinguishable objects into $5$ distinguishable containers. As is explained quite clearly in the article to which I linked, this can be done in $$\binom{5+5-1}{5-1}=\binom94=126$$ ways.

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### Mistu4u

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Updated on June 14, 2022

So, we need to find $a,b,c,d,e\ge 0$ such that $a+b+c+d+e=5$. By divide & conquer approach, I've arrived at $^{5+5-1}C_4=126$ possible values which must be available from some standard formula of which I'm unfortunately not aware.