# Dice sum 3

I have a 2-sided dice with sides labelled $$1,2$$, a 3-sided dice with sides labelled $$1,2,3$$, and so on, up to a 49-sided dice with sides labelled $$1,2,3,...,49$$. I simultaneously roll all 48 of them. Let $$d_{i}$$ denote the number that comes up on the dice with $$i$$ sides. The number of ways that I can roll the dice so that $3 \mid \displaystyle\sum_{i=2}^{49} d_{i}$ can be expressed in the form $$\dfrac{n!}{3}$$. What is $$n$$?

This is the final (and hardest) of the dice sum problems. For the previous problem, see Dice Sums 2.

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