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The number of ways in which an examiner can assign 15 marks to 5 questions giving not less than 2 marks to any question is
a.1 b.126 c.120 d.240
I am getting no clue!

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This problem is very similar: math.stackexchange.com/questions/101432/… – Byron Schmuland Sep 30 '12 at 18:20
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. – lab bhattacharjee Sep 30 '12 at 18:53

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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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