**Question:**

In an examination, a candidate is required to answer 7 questions out of 12, which are divided into two groups, each containing 6 questions. One cannot attempt more than 5 questions from either group. In how many ways can he choose these questions?

**Solution:**

There are total 13 questions out of which 10 is to be answered .The student can answer in the following ways:

$\Rightarrow 3$ questions from part $A$ and 4 from part $B$

$\Rightarrow 4$ questions from part $A$ and 3 from part $B$

$\Rightarrow 5$ questions from part $A$ and 2 from part $B$

$\Rightarrow 2$ questions from part $A$ and 5 from part $B$

$\Rightarrow$ total ways in the 1 st case are ${ }^{6} \mathrm{C}_{3} \times{ }^{6} \mathrm{C}_{4}$

$\Rightarrow$ total ways in the 2 nd case are ${ }^{6} \mathrm{C}_{4} \times{ }^{6} \mathrm{C}_{3}$

$\Rightarrow$ total ways in the 3 rd case are ${ }^{6} \mathrm{C}_{5} \times{ }^{6} \mathrm{C}_{2}$

$\Rightarrow$ total ways in the 4 th case are ${ }^{6} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{5}$

thus the total of the all the cases would be $={ }^{6} \mathrm{C}_{4} \times{ }^{6} \mathrm{C}_{3}+{ }^{6} \mathrm{C}_{3} \times{ }^{6} \mathrm{C}_{4}{ }^{+6}{ }^{6} \mathrm{C}_{5} \times{ }^{6} \mathrm{C}_{2}+{ }^{6} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{5}$

Applying ${ }^{\mathrm{n}} \mathrm{C}_{r}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}$

$=780$ ways.