Solve the following

Given ${ }^{18} C_{15}+2\left({ }^{18} C_{16}\right)+{ }^{17} C_{16}+1={ }^{n} C_{3}$

L.H.S ${ }^{18} \underline{\underline{C}}_{15} \pm \underline{18} \underline{\underline{C}}_{16}+{ }^{18} C_{16}+{ }^{17} C_{16}+1$

Since, ${ }^{n} C_{r+1}-{ }^{n} C_{r}={ }^{n+1} C_{r+1}$ and $1={ }^{17} C_{17}$

L.H.S reduces to, 

${ }^{19} C_{16}+{ }^{18} C_{16}+{ }^{17} \underline{C}_{16}+17 \underline{1} \underline{C}_{17}$

$={ }^{19} C_{16}+1 \underline{18} \underline{C}_{16} \pm \underline{18} \underline{C}_{17}$

= 19C16 + 19C17 

i.e L.H.S = 20C17 and R.H.S = nC3 = L.H.S (given)

∴ 20C17 = nC3 

or 20C3 = nC3    [∵ 20C17 = 20C20−17  = 20C3]

⇒ n = 20