If $\frac{1}{{ }^4{C}_n}=\frac{1}{{ }^5{C}_n}+\frac{1}{{ }^6{C}_n}$,  then n=

$\frac{1}{{ }^4{C}_n}=\frac{1}{{ }^5{C}_n}+\frac{1}{{ }^6{C}_n}$

$\frac{n!(4-n)!}{4!}=\frac{(5-n)!n!}{5!}+\frac{(6-n)!n!}{6!}$

$\frac{(4-n)!}{4!}=\frac{(5-n)(4-n)!}{5\times 4!}+\frac{(6-n)(5-n)(4-n)!}{6\times 5\times 4!}$

$1=\frac{(5-n)}{5}+\frac{(6-n)(5-n)}{6\times 5}$

$1=\left(\frac{30-6n+30-6n-5n+n^2}{30}\right)$

$30=60-17n+n^2$

$n^2-17n+30=0$

$n^2-15n-2n+30=0$

$n(n-15)-2(n-15)=0$

$(n-15)(n-2)=0$

$n=2,15$

$\Rightarrow n=2$

Since  ${}^n{C}_r$ is defined for $n\ge r$