We define a function $f$ on the $m$-tuples $(n_1, n_2, ..., n_m)$ in the following way: $f(n_1, n_2, ..., n_m) = 0$ for the $m$-tuples for which it does not hold that:
$$n_1 \ge n_2 \ge \dots \ge n_m \ge 0. \tag{1}$$
If $n_1 \ge n_2 \ge \dots \ge n_m \ge 0$, then it holds that
$$f(n_1, n_2, \dots, n_m, 0) = f(n_1, n_2, \dots, n_m). \qquad (2)$$
$$f(n_1, n_2, \dots, n_m) = f(n_1 - 1, n_2, \dots, n_m) + f(n_1, n_2 - 1, \dots, n_m) + \dots + f(n_1, n_2, \dots, n_m - 1). \quad (3)$$
$$f(n) = 1, \text{ for } n \ge 0. \qquad (4)$$
The function $f$ is well-defined. We will show that $f(n_1, n_2, \dots, n_m)$ is the number of grids of type $(n_1, n_2, \dots, n_m)$. It is clear that conditions (1) and (4) must be fulfilled for the required grid (a row of $n$ squares can be filled in a unique way with $n$ numbers such that they form an increasing sequence $:1, 2, 3, \dots, n$). Condition (2) is also clear. To show that a grid of type $(n_1, n_2, \dots, n_m)$ satisfies condition (3), we consider the number $n$. That number has to be the last number in one of the rows, so if we delete the square having number $n$, we get a grid with $n-1$ numbers that fulfills the conditions of the problem. It can happen that several rows may have the same number of squares so that the number $n$ can be in the last row in the rightmost square. If, for example, $n_1=n_2$, then $f(n_1-1, n_2, \dots, n_m) = 0$ since $n_1-1 < n_2$. Therefore in condition (3) all expressions
$$f(n_1-1, n_2, \dots, n_m), f(n_1, n_2-1, \dots, n_m), \dots, f(n_1, n_2, \dots, n_m-1)$$
appear, which are zero if the above condition holds.
We use the previously-defined function $f$ to get the number of grids of type $(4,3,2)$.
$$$$
$$\begin{align*} f(4,3,2) &= f(3,3,2) + f(4,2,2) + f(4,3,1) = \\ &= (f(3,2,2) + f(3,3,1)) + (f(3,2,2) + f(4,2,1)) + (f(3,3,1) + f(4,2,1) + f(4,3,0)) = \\ &= 2f(3,2,2) + 2f(3,3,1) + 2f(4,2,1) + f(4,3) = \\ &= 2(f(2,2,2) + f(3,2,1)) + 2(f(3,2,1) + f(3,3,0)) + 2(f(3,2,1) + f(4,1,1) + f(4,2,0)) + (f(3,3) + f(4,2)) = \\ &= 2f(2,2,2) + 6f(3,2,1) + 2f(4,1,1) + 3f(4,2) + 3f(3,3) = \\ &= 2(f(2,2,1)) + 6(f(2,2,1) + f(3,1,1) + f(3,2,0)) + 2(f(3,1,1) + f(4,1,0)) + 3(f(3,2) + f(4,1)) + 3(f(3,2)) = \\ &= 8f(2,2,1) + 8f(3,1,1) + 12f(3,2) + 5f(4,1) = \\ &= 8(f(2,1,1) + f(2,2,0)) + 8(f(2,1,1) + f(3,1,0)) + 12(f(2,2) + f(3,1)) + 5(f(3,1) + f(4,0)) = \\ &= 16f(2,1,1) + 20f(2,2) + 25f(3,1) + 5f(4) = \\ &= 16(f(1,1,1) + f(2,1,0)) + 20(f(2,1)) + 25(f(2,1) + f(3,0)) + 5 = \\ &= 16f(1,1,1) + 61f(2,1) + 25f(3) + 5 = 16(f(1,1,0)) + 61(f(1,1) + f(2,0)) + 25 + 5 = \\ &= 77f(1,1) + 61f(2) + 30 = 77(f(1,0)) + 61 + 30 = 77f(1) + 91 = 77 + 91 = 168. \end{align*}$$