# Dice sum 3

**Discrete Mathematics**Level 4

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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