Japt, 8 bytes

ü ®ÅÊÊÃ×

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Perl 5 -ap, 27 bytes

$=1;$*=$h{$_}++||1for@F}{

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05AB1E, 5 bytes

Ù¢<!P

A bit lame, but this is simply a port of @xnor's Python 2 answer and @Riolku's Jelly answer, so make sure to upvote them!

Try it online or verify all test cases.

Explanation:

Ù      # Uniquify the (implicit) input-list
 ¢     # Count how many times each occurs in the (implicit) input-list
  <    # Decrease each count by 1
   !   # Take the factorial on each value
    P  # And take the product of that
       # (after which it is output implicitly as result)

Jelly, 5 bytes

ĠẈ’!P

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Count Euler's Tours 5 Byte Jelly Solution, as mentioned by HyperNeutrino.

The idea is the same as an earlier python solution, but I'd still like to explain how I came about it.

Note that we can solve this recursively. If our current node is N, with C children, our answer is C! times the product of the sub-answers from all the children. The reason is we can descend (in our euler tour) into the children in any order we so choose.

The second observation is that for every node except the root, where count(x) is the number of occurences of x in the euler tour:

$$count(i) = deg(i)$$

For the root,

$$count(r) = deg(r) + 1$$

Now, if we look at our formula above, we note that our answer is

$$deg(r)! * [deg(n) - 1]!$$

for every other node n. This actually works out hilariously well. We note that, as mentioned before, the answer is:

$$prod_{i = 1}^v [count(v) - 1]!$$

This solution lends itself very well to jelly.

ĠẈ’!P

First, we get the count array that we need, by getting Ġrouping indices by their elements and taking vectorized length. Then we decrement and factorial, and finally product:

ĠẈ      Count array
  ’     Decrement
   !    Factorial
    P   Product

P.S.: Thanks to HyperNeutrino for helping me with the jelly solution, seeing as my jelly is a bit rusty...

Charcoal, 11 bytes

ＩΠＥθ∨№…θκι¹

Try it online! Link is to verbose version of code. Uses @xnor's algorithm. Explanation:

   θ        Input array
  Ｅ         Map over elements
       θ    Input array
      …     Truncated to length
        κ   Current index
     №      Count occurrences of
         ι  Current element
    ∨       Logical Or
          ¹ Literal `1`
 Π          Take the product
Ｉ           Cast to string
            Implicitly print

Using the form of representation whereby leaf nodes are visited once and the root node is considered to have an extra edge so that it gets visited both at the start and the end, then the number of edges for a given node is equal to the number of times it was visited. On the first visit to a node with n edges you only have a choice of n-1 edges since you have to completely traverse all of its subtrees before returning to the parent, and similarly on subsequent visits you have fewer choices until finally the last two visits both give you no choice at all. This could be reduced to the products of the relevant factorials for each node, although as @xnor points out it's actually golfier for some languages to replace each visit with the count of remaining choices and take the overall product (the Charcoal algorithm actually takes the product in reverse order but of course this makes no difference to the total).

JavaScript (ES6), 42 bytes

This is based on @xnor's conjecture.

a=>a.map(x=>t*=(a[~x]=-~a[~x])-1||1,t=1)|t

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Python 2, 45 bytes

f=lambda l:l==[]or(l.count(l.pop())or 1)*f(l)

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It seems that one way to get the answer is take the counts of each node, that is the number of times each node is listed, subtract 1 from each, and multiply their factorials. That is,

$$prod_i (c_i-1)!$$

where node $i$ is listed $c_i$ times. Note that it doesn't matter what order the nodes appear in the Euler Tour Representation, only how many times they do. Listing leaf nodes once or twice doesn't make a difference, since $0!=1!=1$.

Since Python doesn't have a built-in factorial, we make do like this. For each entry of the list, we will count how many times that same values appears before it, that is label its k'th appearance 0-indexed as k. For example,

1 2 4 2 5 2 1 3 6 3 1
0 0 0 1 0 2 1 0 0 1 2  

If we then remove all zeroes (or convert them to 1's) and multiply, we get the desired result.

To do this recursively, we repeatedly remove the last element and consider its its count in the remainder of the list, converting any count of 0 to 1. Then, we multiply by the recursive result on the list with its last element now removed, with a base case of 1 for the empty list.

We can save a couple of bytes using splatted input and True for 1 if this is allowed.

43 bytes

f=lambda a,*l:l==()or(l.count(a)or 1)*f(*l)

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APL (Dyalog Unicode), 17 11 bytes

Based on xnor's answer. The ⍤ operator requires version 18, which is not yet on TIO.

(×/!÷⊢)≢⍤⊢⌸

Here is a longer variant that works in lower versions:

(×/!÷⊢)∘(+/∪∘.=⊢)

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+/∪∘.=⊢/≢⍤⊢⌸ counts the occurrences of the unique items in the input, ×/!÷⊢ calculates $$prod_i{c_i!over c_i}$$

APL (Dyalog Unicode), 27 bytes

I think this works with all suggested input formats.

{2>≢⍵:1⋄(×/∇¨,∘!≢)⍵⊆⍨⍵≠⌊/⍵}

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I don't think the base case 2>≢⍵:1 is really necessary, since at some point there are no subtrees left to recurse on, but I can't get this to work without it.

Commented:

{2>≢⍵:1⋄(×/∇¨,∘!≢)⍵⊆⍨⍵≠⌊/⍵}  ⍝ A recursive dfns
 2>≢⍵                        ⍝ If the input has less than 2 elements
     :1                      ⍝   return 1
                  ⍵⊆⍨        ⍝ Partition the input ...
                     ⍵≠      ⍝ ... taking the elements that are not equal
                       ⌊/⍵   ⍝ to the minimum (the root of the current tree)
                             ⍝ These are the new subtrees
               !≢            ⍝ Take the factorial of the number of subtrees
             ,               ⍝ And append it to the results of ...
           ∇¨                ⍝ the recursive calls on each subtree
         ×/                  ⍝ take the product of the vector
                             ⍝ this is the result