Verify that:

(i) Given: ${ }^{15} \mathrm{C}_{8}+{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{6}-{ }^{15} \mathrm{C}_{7}$

To prove: ${ }^{15} \mathrm{C}_{8}+{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{6}-{ }^{15} \mathrm{C}_{7}=0$

We know that:

${ }^{n} C_{r}={ }^{n} C_{n-r}$

$\Rightarrow{ }^{15} \mathrm{C}_{8}={ }^{15} \mathrm{C}_{7} \&{ }^{15} \mathrm{C}_{9}={ }^{15} \mathrm{C}_{6}$

$\Rightarrow{ }^{15} \mathrm{C}_{8}+{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{6}-{ }^{15} \mathrm{C}_{7}={ }^{15} \mathrm{C}_{8}+{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{8}=0$

Hence, proved that ${ }^{15} \mathrm{C}_{8}+{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{6}-{ }^{15} \mathrm{C}_{7}=0$

(ii) We know that: ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$

Therefore, $n=10$ and $r=4$

L.H.S $={ }^{10} \mathrm{C}_{4}+{ }^{10} \mathrm{C}_{3}={ }^{11} \mathrm{C}_{4}$

Hence, proved.