- Home
- Standard 11
- Mathematics
In an examination of Mathematics paper, there are $20$ questions of equal marks and the question paper is divided into three sections : $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. A student is required to attempt total $15$ questions taking at least $4$ questions from each section. If section $A$ has $8$questions, section $\mathrm{B}$ has $6$ questions and section $\mathrm{C}$ has $6$ questions, then the total number of ways a student can select $15$ questions is
$11370$
$11376$
$11375$
$11350$
Solution
If $4$ questions from each section are selected
Remaining $3$ questions can be selected either in $(1,1,1)$ or $(3,0,0)$ or $(2,1,0)$
$\therefore$ Total ways $={ }^8 \mathrm{c}_5 \cdot{ }^6 \mathrm{c}_5 \cdot{ }^6 \mathrm{c}_5+{ }^8 \mathrm{c}_6{ }^6 \mathrm{c}_5 \cdot{ }^6 \mathrm{c}_4 \times 2+$
${ }^8 \mathrm{c}_5 \cdot{ }^6 \mathrm{c}_6 \cdot{ }^6 \mathrm{c}_4 \times 2+{ }^8 \mathrm{c}_4 \cdot{ }^6 \mathrm{c}_6 \cdot{ }^6 \mathrm{c}_5 \times 2+{ }^8 \mathrm{c}_7 \cdot{ }^6 \mathrm{c}_4 \cdot{ }^6 \mathrm{c}_4 $
$ =56 \cdot 6 \cdot 6+28 \cdot 6 \cdot 15 \cdot 2+56 \cdot 15 \cdot 2+70 \cdot 6 \cdot 2 $
$ +8 \cdot 15 \cdot 15 $
$ =2016+5040+1680+840+1800=11376$
