Verify that: $^{15}C_{8}+^{15}C_{9}-^{15}C_{6}-^{15}C_{7} = 0$

We know that,

${ n }_{ C_{ r } }=\displaystyle \frac { n! }{ r!\left( n-r \right) ! }$

$\therefore LHS={ { 15 }_{ { C }_{ 8 } } }+{ 15 }_{ { C }_{ 9 } }-{ 15 }_{ { C }_{ 6 } }-{ 15 }_{ { C }_{ 7 } }$

$\displaystyle =\dfrac { 15! }{ 8!\left( 15-8 \right) ! } +\frac { 15! }{ 9!\left( 15-9 \right) ! } -\frac { 15! }{ 6!\left( 15-6 \right) ! } -\frac { 15! }{ 7!\left( 15-7 \right) ! }$

$=\displaystyle \dfrac { 15! }{ 8!\times 7! } +\frac { 15! }{ 9!\times 6! } -\frac { 15! }{ 6!\times 9! } -\frac { 15! }{ 7!\times 8! }$

$=\displaystyle \left( \frac { 15! }{ 8!\times 7! } -\frac { 15! }{ 7!\times 8! } \right) +\left( \frac { 15! }{ 9!\times 6! } -\frac { 15! }{ 6!\times 9! } \right)$

$=\left( 0 \right) +\left( 0 \right) =0$

$=RHS$