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

$\begin{array}{1 1}(A)\;72\\(B)\;71\\(C)\;36\\(D)\;\text{None of these}\end{array} $

1 Answer

  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
After giving 1 mark to each question,7 marks to be distributed among 3 question.
The required number of ways =The number of non-negative integral solution of x+y+z=7
$\Rightarrow 9C_7=9C_2$
$9C_2=\large\frac{9!}{2!7!}=\frac{9\times 8\times 7!}{2\times 7!}$
$\Rightarrow 36$
Hence (C) is the correct answer.
answered Jun 20, 2014 by sreemathi.v

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