A standard six-sided die is rolled $6$ times. You are told that among the rolls, there was one $1,$ two $2$'s, and three $3$'s. How many possible sequences of rolls could there have been? (For example, $3,2,3,1,3,2$ is one possible sequence.)

Guest Feb 27, 2021

#1**+1 **

We are finding the number of arrangements of the number "122333".

This is equal to \(\frac{6!}{2! \cdot 3!} = \boxed{60}\)

CubeyThePenguin Feb 27, 2021