How many distinguishable outcomes from rolling 6 identical dice?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Ignoring order, how many distinguishable outcomes are there from rolling 6 identical dice? Answer = $462$
I tried a variety of ways such as $frac6^66!$ and can't seem to get the answer. Struggling how to incorporate no order and distinguishable at the same time. Please help.
combinatorics statistics
edited Feb 16 at 20:57
smci
346211
asked Feb 16 at 12:34
KombatWombatKombatWombat
756
$endgroup$
add a comment |
$begingroup$
Ignoring order, how many distinguishable outcomes are there from rolling 6 identical dice? Answer = $462$
I tried a variety of ways such as $frac6^66!$ and can't seem to get the answer. Struggling how to incorporate no order and distinguishable at the same time. Please help.
combinatorics statistics
edited Feb 16 at 20:57
smci
346211
asked Feb 16 at 12:34
KombatWombatKombatWombat
756
$endgroup$
$begingroup$
Offhand looks like one might need to consider all the possible partitions of 6.
$endgroup$
– coffeemath
Feb 16 at 12:37
add a comment |
$begingroup$
Ignoring order, how many distinguishable outcomes are there from rolling 6 identical dice? Answer = $462$
I tried a variety of ways such as $frac6^66!$ and can't seem to get the answer. Struggling how to incorporate no order and distinguishable at the same time. Please help.
combinatorics statistics
edited Feb 16 at 20:57
smci
346211
asked Feb 16 at 12:34
KombatWombatKombatWombat
756
$endgroup$
Ignoring order, how many distinguishable outcomes are there from rolling 6 identical dice? Answer = $462$
I tried a variety of ways such as $frac6^66!$ and can't seem to get the answer. Struggling how to incorporate no order and distinguishable at the same time. Please help.
combinatorics statistics
combinatorics statistics
edited Feb 16 at 20:57
smci
346211
asked Feb 16 at 12:34
KombatWombatKombatWombat
756
edited Feb 16 at 20:57
smci
346211
asked Feb 16 at 12:34
KombatWombatKombatWombat
756
edited Feb 16 at 20:57
smci
346211
edited Feb 16 at 20:57
smci
346211
edited Feb 16 at 20:57
smci
346211
346211
asked Feb 16 at 12:34
KombatWombatKombatWombat
756
asked Feb 16 at 12:34
KombatWombatKombatWombat
756
asked Feb 16 at 12:34
KombatWombatKombatWombat
756
756
$begingroup$
Offhand looks like one might need to consider all the possible partitions of 6.
$endgroup$
– coffeemath
Feb 16 at 12:37
add a comment |
$begingroup$
Offhand looks like one might need to consider all the possible partitions of 6.
$endgroup$
– coffeemath
Feb 16 at 12:37
$begingroup$
Offhand looks like one might need to consider all the possible partitions of 6.
$endgroup$
– coffeemath
Feb 16 at 12:37
$begingroup$
Offhand looks like one might need to consider all the possible partitions of 6.
$endgroup$
– coffeemath
Feb 16 at 12:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An outcome here is the same as a six-tuple of non-negative integers that sum to $6$, the $i^th$ entry telling you how many times $i$ came up as a value.
Stars and Bars tell us that the number of such is $$binom 6+6-16=462$$
answered Feb 16 at 12:37
lulululu
42.9k25080
$endgroup$
$begingroup$
Could you explain in more detail please, thank you.
$endgroup$
– KombatWombat
Feb 16 at 12:41
1
$begingroup$
What part is confusing? The link contains a detailed proof of the relevant formula.
$endgroup$
– lulu
Feb 16 at 12:45
2
$begingroup$
Is the bijection clear? Since the order of the dice doesn't matter, the only thing that distinguishes two different outcomes is the six-tuple. How many $1's$ did you get? How many $2's$ and so on. Thus you just want to count those six-tuples.
$endgroup$
– lulu
Feb 16 at 12:46
$begingroup$
I'm confused as to why it must sum to 6?
$endgroup$
– KombatWombat
Feb 16 at 12:50
2
$begingroup$
Say you throw and you get the values $1,6,1,1,5,6$ We note that you got three $1's$, one $5$ and two $6's$. Thus your outcome would be the six-tuple $(3,0,0,0,1,2)$. Note that these sum to the total number of dice, namely $6$.
$endgroup$
– lulu
Feb 16 at 12:54
|
show 3 more comments
$begingroup$
Let $x_1,x_2,...,x_6$ indicate the number of $1,2,3,4,5,6$.
Then the problem can be formulated as:
$$x_1+x_2+x_3+x_4+x_5+x_6=6, 0le x_ile 6.$$
For example, the following outcomes are equivalent:
$$111112equiv 111121equiv 111211equiv 112111equiv 121111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(5,1,0,0,0,0);\
111123equiv 111213equiv 112113equiv 121113equiv cdotsequiv 321111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(4,1,1,0,0,0);\$$
Using Stars and Bars method:
$$6+6-1choose 6-1=462.$$
answered Feb 16 at 12:56
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An outcome here is the same as a six-tuple of non-negative integers that sum to $6$, the $i^th$ entry telling you how many times $i$ came up as a value.
Stars and Bars tell us that the number of such is $$binom 6+6-16=462$$
answered Feb 16 at 12:37
lulululu
42.9k25080
$endgroup$
$begingroup$
Could you explain in more detail please, thank you.
$endgroup$
– KombatWombat
Feb 16 at 12:41
1
$begingroup$
What part is confusing? The link contains a detailed proof of the relevant formula.
$endgroup$
– lulu
Feb 16 at 12:45
2
$begingroup$
Is the bijection clear? Since the order of the dice doesn't matter, the only thing that distinguishes two different outcomes is the six-tuple. How many $1's$ did you get? How many $2's$ and so on. Thus you just want to count those six-tuples.
$endgroup$
– lulu
Feb 16 at 12:46
$begingroup$
I'm confused as to why it must sum to 6?
$endgroup$
– KombatWombat
Feb 16 at 12:50
2
$begingroup$
Say you throw and you get the values $1,6,1,1,5,6$ We note that you got three $1's$, one $5$ and two $6's$. Thus your outcome would be the six-tuple $(3,0,0,0,1,2)$. Note that these sum to the total number of dice, namely $6$.
$endgroup$
– lulu
Feb 16 at 12:54
|
show 3 more comments
$begingroup$
An outcome here is the same as a six-tuple of non-negative integers that sum to $6$, the $i^th$ entry telling you how many times $i$ came up as a value.
Stars and Bars tell us that the number of such is $$binom 6+6-16=462$$
answered Feb 16 at 12:37
lulululu
42.9k25080
$endgroup$
$begingroup$
Could you explain in more detail please, thank you.
$endgroup$
– KombatWombat
Feb 16 at 12:41
1
$begingroup$
What part is confusing? The link contains a detailed proof of the relevant formula.
$endgroup$
– lulu
Feb 16 at 12:45
2
$begingroup$
Is the bijection clear? Since the order of the dice doesn't matter, the only thing that distinguishes two different outcomes is the six-tuple. How many $1's$ did you get? How many $2's$ and so on. Thus you just want to count those six-tuples.
$endgroup$
– lulu
Feb 16 at 12:46
$begingroup$
I'm confused as to why it must sum to 6?
$endgroup$
– KombatWombat
Feb 16 at 12:50
2
$begingroup$
Say you throw and you get the values $1,6,1,1,5,6$ We note that you got three $1's$, one $5$ and two $6's$. Thus your outcome would be the six-tuple $(3,0,0,0,1,2)$. Note that these sum to the total number of dice, namely $6$.
$endgroup$
– lulu
Feb 16 at 12:54
|
show 3 more comments
$begingroup$
An outcome here is the same as a six-tuple of non-negative integers that sum to $6$, the $i^th$ entry telling you how many times $i$ came up as a value.
Stars and Bars tell us that the number of such is $$binom 6+6-16=462$$
answered Feb 16 at 12:37
lulululu
42.9k25080
$endgroup$
An outcome here is the same as a six-tuple of non-negative integers that sum to $6$, the $i^th$ entry telling you how many times $i$ came up as a value.
Stars and Bars tell us that the number of such is $$binom 6+6-16=462$$
answered Feb 16 at 12:37
lulululu
42.9k25080
answered Feb 16 at 12:37
lulululu
42.9k25080
answered Feb 16 at 12:37
lulululu
42.9k25080
answered Feb 16 at 12:37
lulululu
42.9k25080
42.9k25080
$begingroup$
Could you explain in more detail please, thank you.
$endgroup$
– KombatWombat
Feb 16 at 12:41
1
$begingroup$
What part is confusing? The link contains a detailed proof of the relevant formula.
$endgroup$
– lulu
Feb 16 at 12:45
2
$begingroup$
Is the bijection clear? Since the order of the dice doesn't matter, the only thing that distinguishes two different outcomes is the six-tuple. How many $1's$ did you get? How many $2's$ and so on. Thus you just want to count those six-tuples.
$endgroup$
– lulu
Feb 16 at 12:46
$begingroup$
I'm confused as to why it must sum to 6?
$endgroup$
– KombatWombat
Feb 16 at 12:50
2
$begingroup$
Say you throw and you get the values $1,6,1,1,5,6$ We note that you got three $1's$, one $5$ and two $6's$. Thus your outcome would be the six-tuple $(3,0,0,0,1,2)$. Note that these sum to the total number of dice, namely $6$.
$endgroup$
– lulu
Feb 16 at 12:54
|
show 3 more comments
$begingroup$
Could you explain in more detail please, thank you.
$endgroup$
– KombatWombat
Feb 16 at 12:41
1
$begingroup$
What part is confusing? The link contains a detailed proof of the relevant formula.
$endgroup$
– lulu
Feb 16 at 12:45
2
$begingroup$
Is the bijection clear? Since the order of the dice doesn't matter, the only thing that distinguishes two different outcomes is the six-tuple. How many $1's$ did you get? How many $2's$ and so on. Thus you just want to count those six-tuples.
$endgroup$
– lulu
Feb 16 at 12:46
$begingroup$
I'm confused as to why it must sum to 6?
$endgroup$
– KombatWombat
Feb 16 at 12:50
2
$begingroup$
Say you throw and you get the values $1,6,1,1,5,6$ We note that you got three $1's$, one $5$ and two $6's$. Thus your outcome would be the six-tuple $(3,0,0,0,1,2)$. Note that these sum to the total number of dice, namely $6$.
$endgroup$
– lulu
Feb 16 at 12:54
$begingroup$
Could you explain in more detail please, thank you.
$endgroup$
– KombatWombat
Feb 16 at 12:41
$begingroup$
Could you explain in more detail please, thank you.
$endgroup$
– KombatWombat
Feb 16 at 12:41
1
1
$begingroup$
What part is confusing? The link contains a detailed proof of the relevant formula.
$endgroup$
– lulu
Feb 16 at 12:45
$begingroup$
What part is confusing? The link contains a detailed proof of the relevant formula.
$endgroup$
– lulu
Feb 16 at 12:45
2
2
$begingroup$
Is the bijection clear? Since the order of the dice doesn't matter, the only thing that distinguishes two different outcomes is the six-tuple. How many $1's$ did you get? How many $2's$ and so on. Thus you just want to count those six-tuples.
$endgroup$
– lulu
Feb 16 at 12:46
$begingroup$
Is the bijection clear? Since the order of the dice doesn't matter, the only thing that distinguishes two different outcomes is the six-tuple. How many $1's$ did you get? How many $2's$ and so on. Thus you just want to count those six-tuples.
$endgroup$
– lulu
Feb 16 at 12:46
$begingroup$
I'm confused as to why it must sum to 6?
$endgroup$
– KombatWombat
Feb 16 at 12:50
$begingroup$
I'm confused as to why it must sum to 6?
$endgroup$
– KombatWombat
Feb 16 at 12:50
2
2
$begingroup$
Say you throw and you get the values $1,6,1,1,5,6$ We note that you got three $1's$, one $5$ and two $6's$. Thus your outcome would be the six-tuple $(3,0,0,0,1,2)$. Note that these sum to the total number of dice, namely $6$.
$endgroup$
– lulu
Feb 16 at 12:54
$begingroup$
Say you throw and you get the values $1,6,1,1,5,6$ We note that you got three $1's$, one $5$ and two $6's$. Thus your outcome would be the six-tuple $(3,0,0,0,1,2)$. Note that these sum to the total number of dice, namely $6$.
$endgroup$
– lulu
Feb 16 at 12:54
|
show 3 more comments
$begingroup$
Let $x_1,x_2,...,x_6$ indicate the number of $1,2,3,4,5,6$.
Then the problem can be formulated as:
$$x_1+x_2+x_3+x_4+x_5+x_6=6, 0le x_ile 6.$$
For example, the following outcomes are equivalent:
$$111112equiv 111121equiv 111211equiv 112111equiv 121111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(5,1,0,0,0,0);\
111123equiv 111213equiv 112113equiv 121113equiv cdotsequiv 321111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(4,1,1,0,0,0);\$$
Using Stars and Bars method:
$$6+6-1choose 6-1=462.$$
answered Feb 16 at 12:56
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
$begingroup$
Let $x_1,x_2,...,x_6$ indicate the number of $1,2,3,4,5,6$.
Then the problem can be formulated as:
$$x_1+x_2+x_3+x_4+x_5+x_6=6, 0le x_ile 6.$$
For example, the following outcomes are equivalent:
$$111112equiv 111121equiv 111211equiv 112111equiv 121111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(5,1,0,0,0,0);\
111123equiv 111213equiv 112113equiv 121113equiv cdotsequiv 321111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(4,1,1,0,0,0);\$$
Using Stars and Bars method:
$$6+6-1choose 6-1=462.$$
answered Feb 16 at 12:56
farruhotafarruhota
20.8k2741
$endgroup$
add a comment |
$begingroup$
Let $x_1,x_2,...,x_6$ indicate the number of $1,2,3,4,5,6$.
Then the problem can be formulated as:
$$x_1+x_2+x_3+x_4+x_5+x_6=6, 0le x_ile 6.$$
For example, the following outcomes are equivalent:
$$111112equiv 111121equiv 111211equiv 112111equiv 121111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(5,1,0,0,0,0);\
111123equiv 111213equiv 112113equiv 121113equiv cdotsequiv 321111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(4,1,1,0,0,0);\$$
Using Stars and Bars method:
$$6+6-1choose 6-1=462.$$
answered Feb 16 at 12:56
farruhotafarruhota
20.8k2741
$endgroup$
Let $x_1,x_2,...,x_6$ indicate the number of $1,2,3,4,5,6$.
Then the problem can be formulated as:
$$x_1+x_2+x_3+x_4+x_5+x_6=6, 0le x_ile 6.$$
For example, the following outcomes are equivalent:
$$111112equiv 111121equiv 111211equiv 112111equiv 121111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(5,1,0,0,0,0);\
111123equiv 111213equiv 112113equiv 121113equiv cdotsequiv 321111 Rightarrow \
(x_1,x_2,x_3,x_4,x_5,x_6)=(4,1,1,0,0,0);\$$
Using Stars and Bars method:
$$6+6-1choose 6-1=462.$$
answered Feb 16 at 12:56
farruhotafarruhota
20.8k2741
answered Feb 16 at 12:56
farruhotafarruhota
20.8k2741
answered Feb 16 at 12:56
farruhotafarruhota
20.8k2741
answered Feb 16 at 12:56
farruhotafarruhota
20.8k2741
20.8k2741
add a comment |
add a comment |
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$begingroup$
Offhand looks like one might need to consider all the possible partitions of 6.
$endgroup$
– coffeemath
Feb 16 at 12:37