If $^{n}C_{14} = ^{n}C_{16},$ find $^{n}C_{28}$.

Given, ${ n }_{ { C }_{ 14 } }={ n }_{ { C }_{ 16 } }$

$\therefore \dfrac { n! }{ 14!\left( n-14 \right) ! } =\dfrac { n! }{ 16!\left( n-16 \right) ! }$

$\therefore \dfrac { 1 }{ 14!\left( n-14 \right) \times \left( n-15 \right) \times \left( n-16 \right) ! } =\dfrac { 1 }{ 16\times 15\times 14!\left( n-16 \right) ! }$

$\displaystyle \therefore \frac { 1 }{ \left( n-14 \right) \times \left( n-15 \right) } =\frac { 1 }{ 16\times 15 }$

$\displaystyle \therefore \left( n-14 \right) \times \left( n-15 \right) =16\times 15$

$\displaystyle \therefore { n }^{ 2 }-29n+210=240$

$\displaystyle \therefore { n }^{ 2 }-29n-30=0$

$\displaystyle \therefore { n }^{ 2 }-30n+n-30=0$

$\displaystyle \therefore n\left( n-30 \right) +1\left( n-30 \right) =0$

$\displaystyle \therefore \left( n+1 \right) \left( n-30 \right) =0$

$\therefore n=-1$ or $n=30$

Consider only positive root

$\therefore n=30$

${ n }_{ { C }_{ 28 } }={ 30 }_{ { C }_{ 28 } }$

$\displaystyle \therefore { 30 }_{ { C }_{ 28 } }=\frac { 30! }{ 28!\left( 30-28 \right) ! }$

$\displaystyle \therefore { 30 }_{ { C }_{ 28 } }=\frac { 30\times 29\times 28! }{ 28!\times 2! }$

$\displaystyle \therefore { 30 }_{ { C }_{ 28 } }=\frac { 30\times 29 }{ 2 }$

$\therefore { 30 }_{ { C }_{ 28 } }=15\times 29$

$\therefore { 30 }_{ { C }_{ 28 } }=435$