Consider the experiment in which five six-sided dice are thrown simultaneously.
How many distinguishable patterns are there, if we are interested purely in the
tally of the occurrences of each number?
Hello, matty888!
Consider the experiment in which five 6-sided dice are thrown simultaneously.
How many distinguishable patterns are there, if we are interested purely
in the tally of the occurrences of each number?
I will assume that this means: the order of the numbers is not considered,
. . similar to evaluating a Poker hand.
Then there are seven possible "hands" . . .
Five-of-a-Kind
There are 6 possible values for the Five-of-a-Kind, and one way to get each of them.
. . There are: .6 Five-of-a-Kind hands.
Four-of-a-Kind
There are 6 possible values for the Four-of-a-Kind.
There are 5 possible values fo the fifth die.
. . There are: .$\displaystyle 6\cdot5 \:=\:{\color{blue}30}$ Four-of-a-Kind hands.
Full House (Three-of-a-kind and a Pair)
There are 6 possible values for the Three-of-a-kind.
There are 5 possible values for the Pair.
. . There are: .$\displaystyle 6\cdot5 \:=\:{\color{blue}30}$ Full Houses.
Three-of-a-Kind
There are 6 possible values for the Three-of-a-Kind.
There are: $\displaystyle {5\choose2} = 10$ choices for the other two dice.
. . There are: .$\displaystyle 6\cdot10 \:=\:{\color{blue}60}$ Three-of-a-Kind hands.
Two Pairs
There are: $\displaystyle {6\choose2} = 15$ choices for the values of the Pairs.
There are: 4 choices for the value of the fifth die.
. . There are: .$\displaystyle 15\cdot4 \:=\:{\color{blue}60}$ Two-Pair hands.
One Pair
There are 6 possible values for the one Pair.
There are: $\displaystyle {5\choose3} = 10$ choices for the other three dice.
. . There are: .$\displaystyle 6\cdot 10 \:=\:{\color{blue}60}$ One-Pair hands.
No Pairs
There are $\displaystyle {6\choose5} = {\color{blue}6}$ No-Pair hands.
Therefore, there are: .$\displaystyle 6 + 30 + 30 + 60 + 60 + 60 + 6 \;=\;\boxed{252}$ possible patterns ("hands").