**Question:**

In an examination, a student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can these questions be chosen?

**Solution:**

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

$\Rightarrow 6$ questions from part $A$ and 4 from part $B$

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

$\Rightarrow 4$ questions from part $A$ and 6 from part $B$

$\Rightarrow$ total ways in the 1 st case are ${ }^{6} \mathrm{C}_{6} \times{ }^{7} \mathrm{C}_{4}$

$\Rightarrow$ total ways in the 2 nd case are ${ }^{6} \mathrm{C}_{5} \times{ }^{7} \mathrm{C}_{5}$

$\Rightarrow$ total ways in the 3 rd case are ${ }^{6} \mathrm{C}_{4}{ }^{\times}{ }^{7} \mathrm{C}_{6}$

thus the total of the all the cases would be total ways in the 1st case is $={ }^{6} \mathrm{C}_{6} \times{ }^{7} \mathrm{C}_{4}+$ ${ }^{6} \mathrm{C}_{5} \times{ }^{\times}{ }^{7} \mathrm{C}_{5}+{ }^{+} \mathrm{C}_{4} \times{ }^{7} \mathrm{C}_{6}$

Applying ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$

$=266$ ways.