There are \(\binom{6}{5}\) ways to choose a roll containing 5 distinct numbers and \(5!\) ways to order those rolls. This is taken out of \(6^5\) possible rolls for five dice.

There are \(\binom{6}{1}\) choices for which number has a pairing and \(\binom{5}{2}\) ways to choose that pair from a roll of 5 dice. There are \(\binom{6-1}{3}\) ways to choose the remaining values (one value from the original 6 is off-limits) with a total of \(3!\) orderings, again taken out of \(6^5\) possible events.

There are \(\binom{6}{2}\) ways to choose the two numbers that make up each pair. From here, we have \(\binom{5}{2}\) ways to choose the first pair (choosing 2 dice from 5) and \(\binom{3}{2}\) ways to choose the second (choosing another 2 dice from remaining 3).

From here our only remaining choice is the value of the final die, choosing 1 value from 4 remaining options, or \(\binom{4}{1}\).

There are \(\binom{6}{1}\) ways to choose which number makes up our three-alike and \(\binom{5}{3}\) ways to select three dice from five. \(2!\binom{6-1}{2}\) accounts for the selection of the remaining two from five potential values with \(2!\) potential orderings.

There are \(\binom{6}{1}\) ways to select which value has a three-alike pairing and \(\binom{5}{3}\) ways to select that pairing from five dice. Since the remaining two dice must have the same value and one value from the original 6 is off-limits, we're selecting one value from the remaining five, or \(\binom{6-1}{1}\).

There are again \(\binom{6}{1}\) ways to decide which value will make up our four-alike and \(\binom{5}{4}\) ways to select four dice from five. \(\binom{6-1}{1}\) accounts for the ways to select a value for the remaining die.

Given \(\binom{6}{1}\) ways to select which value has a five-alike pairing, there's only \(\binom{5}{5} = 1\) way to roll each value.