Question

The number of different ways of distributing 10 marks among 3 questions, each carrying at least 1 mark, is

• A. $72$
• B. $71$
• C. $36$
• D. none of these

Answer 1

The correct option is C $36$

Consider 3 questions
$ABC$
If $A=1mark$
Then
9 marks can be distributed among B and C in 8 different ways, since we do not consider the possibility of zero marks. That is both B and C holds at-least 1 mark.
Hence for $A=1marks$ we have 8 ways of distributing 10 marks among 3 questions.
Similarly
For $A=2marks$ we will have 7 ways of distributing remaining 8 marks among B and C, Hence there will be 7 ways of distributing 10 marks amongst 3 questions.
For $A=3$ we will have 6 ways of distributing 10 marks among 3 questions and so on
:
:
Hence for $A=9$ marks we will have only one way of distributing 10 marks among 3 question A, B,C.
Thus the total number of ways of distributing 10 marks among 3 questions such that each has at-least 1 marks is
$=8+7+6+5+...+1$
$=$ Sum of first 8 natural numbers.
$=\frac{8(8+1)}{2}$
$=\frac{72}{2}$
$=36$