Combinatorics monster???

Let $N = 2015$ for convenience.

Since the order of M and T among themselves is fixed, we lump them together as "X". Thus we must find the number of possible arrangements of $\text{X}^{4N}\ \ \text{A}^{2N}\ \ \text{H}^{N}\ \ \text{E}^{N}\ \ \text{I}^{N}\ \ \text{C}^{N}\ \ \text{S}^{N}.$ The ways in which this can be done is $\frac{(4N+2N+N+N+N+N+N)!}{(4N)!\ (2N)!\ N!\ N!\ N!\ N!\ N!} = \frac{(11N)!}{(4N)!\ (2N)!\ (N!)^5};$ however, since the first of the "X"s may be chosen to be either M or T, we must double this, so that the number of words is $2\cdot\frac{22165!}{8060!\cdot 4030!\cdot (2015!)^5}.$ Add: $22165 + 8060 + 4030 + 2 + 2015 = 36272.$

$Let's\quad Generalize\quad for\quad letters\quad written\quad n\quad times\\ Let\quad M\quad and\quad T\quad be\quad placed\quad in\quad order\\ \quad M\quad \quad T\quad \quad M\quad \quad T\quad \quad M\quad \quad T\quad \quad M\quad \quad T\quad ...\quad 4n\quad letters\quad \cdots \quad 1\\ \uparrow \quad \quad \uparrow \quad \quad \uparrow \quad \quad \uparrow \quad \quad \uparrow \quad \quad \uparrow \quad \quad \uparrow \quad \quad \uparrow \quad ...\quad 4n+1\quad places\\ \quad (to\quad place\quad rest\quad letters)\\ let\quad spaces\quad be\quad { x }_{ i }\quad 1\le i\le 4n+1\\ =>\quad { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }+...=7n\quad (n*(11-4))\quad \cdots \quad 2\\ { x }_{ i }\ge 0\\ no.\quad of\quad solutions\quad =\quad coefficient\quad of\quad { x }^{ 7n }\quad in\quad { (1+x+{ x }^{ 2 }+...) }^{ 4n+1 }\\ \quad \qquad \qquad \qquad \qquad \quad =\quad coefficient\quad of\quad { x }^{ 7n }\quad in\quad { (1-x) }^{ -(4n+1) }\\ \qquad \qquad \qquad \qquad \quad \quad =\quad (\begin{matrix} 11n \\ 7n \end{matrix})\\ Now\quad including\quad rearrangements\quad in\quad 1\quad and\quad 2\\ \frac { (11n)! }{ (7n)!(4n)! } \cdot \frac { (7n)! }{ (2n)!{ (n!) }^{ 5 } } \cdot 2\quad =\quad 2\times \frac { (11n)! }{ (4n)!(2n)!{ (n!) }^{ 5 } } \\ plug\quad in\quad n=2015\quad and\quad get\quad answer$