Let $( n _ { 1 } , n _ { 2 } , \cdots , n _ { 12 } )$ be a permutation of the numbers $1,2 , \cdots , 12$. The number of arrangements with
$$n _ { 1 } > n _ { 2 } > n _ { 3 } > n _ { 4 } > n _ { 5 } > n _ { 6 }$$
and
$$n _ { 6 } < n _ { 7 } < n _ { 8 } < n _ { 9 } < n _ { 10 } < n _ { 11 } < n _ { 12 }$$
equals:\\
(A) $\binom { 12 } { 5 }$\\
(B) $\binom { 12 } { 6 }$\\
(C) $\binom { 11 } { 6 }$\\
(D) $\frac { 11 ! } { 2 }$