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A student is to answer $10$ questions out of $13$ in an examination, so that he must choose at least $4$ from first $5$ questions. The no. of choices available to him is ?

$\begin{array}{1 1} 140 \\ 196 \\ 280 \\ 346 \end{array}$

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Toolbox:
  • If the conjunction is 'and' between the selections then do multiplication and if the conjunction is 'either or'.then add.
The required choice can be done by:
either (4 from $1^{st}$ 5 and 6 from remaining 8) or (5 from $1^{st}$ 5 and 5 from remaining 8)
$4$ questions out of $5$ can be selected in $^5C_4=5$ ways
$6$ questions from the remaining $8$ can be selected in $^8C_6=28$ ways.
5 from 5 questions can be selected in $^5C_5=1$ way
5 from remaining 8 questions can be selected in $^8C_5=56$ ways.
$\therefore\:$ Total no. of choices $=(5\times 28)+(1\times 56)=140+56=196$

 

answered Aug 8, 2013 by rvidyagovindarajan_1
edited Aug 8, 2013 by rvidyagovindarajan_1
 

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