Lower bound for merging $m$ sorted arrays (decision tree leaves count - permutations) I need some help understanding how to calculate the lower bound on the time complexity of merging $m$ sorted arrays of length $n$.

The bound should be $nm \lg(m)$. I need to prove this using a decision tree.

I tried counting the number of possible permutations (which would be the number of leaves in the tree) but 
I got stuck. I tried the following:

When merging the 2nd array with the first, we have $\binom{2n}{n}$ possibilities to place the elements. When merging the 3rd, we have $\binom{3n}{n}$ and so on, so the total number of permutations is:

$$\binom{2n}{n} \binom{3n}{n} \dots \binom{mn}{n}$$

The height of the tree is denoted $h$.

$$h &gt; \lg{\binom{2n}{n} \binom{3n}{n} \dots \binom{mn}{n}}$$

And from here I don't know how to prove the complexity.

My question is - did I count the permutations correctly? Is there a more simple bound? How to continue from here?

Also, ould this problem possibly be equivalent to the problem of sorting an $m$-sorted array of size $nm$? Because in this problem it is easy to prove that the bound is $nm \lg(m)$.
Stirling's formula shows that
$$
\binom{N}{pN} \sim \frac{2^{H(p)N}}{\sqrt{2\pi p(1-p)N}},
$$
where $H(p)$ is the binary entropy function:
$$
H(p) = p\log \frac{1}{p} + (1-p) \log \frac{1}{1-p}.
$$
In particular, this shows that for $k \geq 2$,
$$
\binom{kn}{n} = (1+o(1)) \frac{2^{H(1/k)kn}}{\sqrt{2\pi(1/k)(1-1/k)kn}} \geq (1+o(1)) \frac{2^{n\log k}}{\sqrt{2\pi n}}.
$$
Therefore
$$
\log \binom{kn}{n} \geq n\log k - O(\log n),
$$
where the big O doesn't depend on $k$. Summing this for $k=2,\ldots,m$, we deduce
$$
h \geq n \log m! - O(m\log n) \geq nm \log m - O(nm),
$$
using Stirling's formula again.