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All combinations of three different dices of side $4, 5$ and $6$.

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0

$begingroup$

You have three dice each with a different number of sides, and numbers ranging from $1$ to the number of sides.

$d_1 = 1 sim 4$

$d_2 = 1 sim 5$

$d_3 = 1 sim 6$

How many possible combinations are out there rolling these three dice?
($[1,1,2]$ will be the same as $[1,2,1]$ and $[2,1,1]$)

I tried to hard figuring this out with combinations but I can't get my head around it.

combinatorics combinations

share|cite|improve this question

edited Mar 16 at 11:11

Vinyl_cape_jawa

3,34011433

asked Mar 16 at 10:15

LeonardLeonard

1033

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Please read this text about how to ask a good question.
  $endgroup$
  – José Carlos Santos
  Mar 16 at 10:22
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What did you "try hard" to do? Where exactly do you need our help?
  $endgroup$
  – Oscar Lanzi
  Mar 16 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I had a wrong approach to begin with. It seemed misleading so I decided not to include.
  $endgroup$
  – Leonard
  Mar 16 at 15:48
  

  
  
  
  

add a comment | 

0

$begingroup$

You have three dice each with a different number of sides, and numbers ranging from $1$ to the number of sides.

$d_1 = 1 sim 4$

$d_2 = 1 sim 5$

$d_3 = 1 sim 6$

How many possible combinations are out there rolling these three dice?
($[1,1,2]$ will be the same as $[1,2,1]$ and $[2,1,1]$)

I tried to hard figuring this out with combinations but I can't get my head around it.

combinatorics combinations

share|cite|improve this question

edited Mar 16 at 11:11

Vinyl_cape_jawa

3,34011433

asked Mar 16 at 10:15

LeonardLeonard

1033

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Please read this text about how to ask a good question.
  $endgroup$
  – José Carlos Santos
  Mar 16 at 10:22
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What did you "try hard" to do? Where exactly do you need our help?
  $endgroup$
  – Oscar Lanzi
  Mar 16 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I had a wrong approach to begin with. It seemed misleading so I decided not to include.
  $endgroup$
  – Leonard
  Mar 16 at 15:48
  

  
  
  
  

add a comment | 

$begingroup$

You have three dice each with a different number of sides, and numbers ranging from $1$ to the number of sides.

$d_1 = 1 sim 4$

$d_2 = 1 sim 5$

$d_3 = 1 sim 6$

How many possible combinations are out there rolling these three dice?
($[1,1,2]$ will be the same as $[1,2,1]$ and $[2,1,1]$)

I tried to hard figuring this out with combinations but I can't get my head around it.

combinatorics combinations

share|cite|improve this question

edited Mar 16 at 11:11

Vinyl_cape_jawa

3,34011433

asked Mar 16 at 10:15

LeonardLeonard

1033

$endgroup$

share|cite|improve this question

edited Mar 16 at 11:11

Vinyl_cape_jawa

3,34011433

asked Mar 16 at 10:15

LeonardLeonard

1033

share|cite|improve this question

edited Mar 16 at 11:11

Vinyl_cape_jawa

3,34011433

asked Mar 16 at 10:15

LeonardLeonard

1033

edited Mar 16 at 11:11

Vinyl_cape_jawa

3,34011433

edited Mar 16 at 11:11

Vinyl_cape_jawa

3,34011433

asked Mar 16 at 10:15

LeonardLeonard

1033

asked Mar 16 at 10:15

LeonardLeonard

1033

2 Answers
2

active

oldest

votes

4

$begingroup$

First, let's figure out how many combinations we can have with three $6$-sided dice. It's the same as counting the number of solutions of the equation
$$x_1+x_2+x_3+x_4+x_5+x_6=3$$
in nonnegative integers; by the so-called "stars and bars" formula, the answer is $binom83=56$.

Not all of these $56$ combinations are possible with your actual set of dice, some of them have too many $5$'s and/or $6$'s; but the number of impossible combinations is small enough to count by hand.

Three sixes: $[6,6,6]$

Two sixes: $[6,6,1]$, $[6,6,2]$, $[6,6,3]$, $[6,6,4]$, $[6,6,5]$

One six and two fives: $[6,5,5]$

Three fives: $[5,5,5]$

That's $8$ impossible combinations, so the answer to your question is $56-8=48$.

share|cite|improve this answer

answered Mar 16 at 12:12

bofbof

52.5k558121

$endgroup$

add a comment | 

1

$begingroup$

Let list all possible combinations together with their cardinality ($xyz$ mean numbers less than or equal to 4):
$$beginalign
&[x56]:&4\
&[xy6]:&4+frac122=10\
&[x55]:&4\
&[xy5]+[x5y]:&4+frac122=10\
&[xyz]:& 4+frac363+frac246=20\
& textall:&48
endalign
$$

Explanation:

$4$ stays for possible combinations with $xy$ or $xyz$ being equal.

$12=4cdot3$ stays for possible combinations with distinct $xy$. It should be divided by $2=frac2!1!1!$ to avoid double counting.

$36=binom31 4cdot3 $ stays for possible combinations with two of $xyz$ being equal and distinct from the third one. It should be divided by $3=frac3!1!2!$ to avoid double counting.

$24=4cdot3cdot2 $ stays for possible combinations with all three of $xyz$ being distinct. It should be divided by $6=frac3!1!1!1!$ to avoid double counting.

share|cite|improve this answer

edited Mar 16 at 11:37

answered Mar 16 at 10:45

useruser

5,81311031

$endgroup$

add a comment | 

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4

$begingroup$

First, let's figure out how many combinations we can have with three $6$-sided dice. It's the same as counting the number of solutions of the equation
$$x_1+x_2+x_3+x_4+x_5+x_6=3$$
in nonnegative integers; by the so-called "stars and bars" formula, the answer is $binom83=56$.

Not all of these $56$ combinations are possible with your actual set of dice, some of them have too many $5$'s and/or $6$'s; but the number of impossible combinations is small enough to count by hand.

Three sixes: $[6,6,6]$

Two sixes: $[6,6,1]$, $[6,6,2]$, $[6,6,3]$, $[6,6,4]$, $[6,6,5]$

One six and two fives: $[6,5,5]$

Three fives: $[5,5,5]$

That's $8$ impossible combinations, so the answer to your question is $56-8=48$.

share|cite|improve this answer

answered Mar 16 at 12:12

bofbof

52.5k558121

$endgroup$

add a comment | 

4

$begingroup$

First, let's figure out how many combinations we can have with three $6$-sided dice. It's the same as counting the number of solutions of the equation
$$x_1+x_2+x_3+x_4+x_5+x_6=3$$
in nonnegative integers; by the so-called "stars and bars" formula, the answer is $binom83=56$.

Not all of these $56$ combinations are possible with your actual set of dice, some of them have too many $5$'s and/or $6$'s; but the number of impossible combinations is small enough to count by hand.

Three sixes: $[6,6,6]$

Two sixes: $[6,6,1]$, $[6,6,2]$, $[6,6,3]$, $[6,6,4]$, $[6,6,5]$

One six and two fives: $[6,5,5]$

Three fives: $[5,5,5]$

That's $8$ impossible combinations, so the answer to your question is $56-8=48$.

share|cite|improve this answer

answered Mar 16 at 12:12

bofbof

52.5k558121

$endgroup$

add a comment | 

$begingroup$

First, let's figure out how many combinations we can have with three $6$-sided dice. It's the same as counting the number of solutions of the equation
$$x_1+x_2+x_3+x_4+x_5+x_6=3$$
in nonnegative integers; by the so-called "stars and bars" formula, the answer is $binom83=56$.

Not all of these $56$ combinations are possible with your actual set of dice, some of them have too many $5$'s and/or $6$'s; but the number of impossible combinations is small enough to count by hand.

Three sixes: $[6,6,6]$

Two sixes: $[6,6,1]$, $[6,6,2]$, $[6,6,3]$, $[6,6,4]$, $[6,6,5]$

One six and two fives: $[6,5,5]$

Three fives: $[5,5,5]$

That's $8$ impossible combinations, so the answer to your question is $56-8=48$.

share|cite|improve this answer

answered Mar 16 at 12:12

bofbof

52.5k558121

$endgroup$

share|cite|improve this answer

answered Mar 16 at 12:12

bofbof

52.5k558121

answered Mar 16 at 12:12

bofbof

52.5k558121

answered Mar 16 at 12:12

bofbof

52.5k558121

1

$begingroup$

Let list all possible combinations together with their cardinality ($xyz$ mean numbers less than or equal to 4):
$$beginalign
&[x56]:&4\
&[xy6]:&4+frac122=10\
&[x55]:&4\
&[xy5]+[x5y]:&4+frac122=10\
&[xyz]:& 4+frac363+frac246=20\
& textall:&48
endalign
$$

Explanation:

$4$ stays for possible combinations with $xy$ or $xyz$ being equal.

$12=4cdot3$ stays for possible combinations with distinct $xy$. It should be divided by $2=frac2!1!1!$ to avoid double counting.

$36=binom31 4cdot3 $ stays for possible combinations with two of $xyz$ being equal and distinct from the third one. It should be divided by $3=frac3!1!2!$ to avoid double counting.

$24=4cdot3cdot2 $ stays for possible combinations with all three of $xyz$ being distinct. It should be divided by $6=frac3!1!1!1!$ to avoid double counting.

share|cite|improve this answer

edited Mar 16 at 11:37

answered Mar 16 at 10:45

useruser

5,81311031

$endgroup$

add a comment | 

1

$begingroup$

Let list all possible combinations together with their cardinality ($xyz$ mean numbers less than or equal to 4):
$$beginalign
&[x56]:&4\
&[xy6]:&4+frac122=10\
&[x55]:&4\
&[xy5]+[x5y]:&4+frac122=10\
&[xyz]:& 4+frac363+frac246=20\
& textall:&48
endalign
$$

Explanation:

$4$ stays for possible combinations with $xy$ or $xyz$ being equal.

$12=4cdot3$ stays for possible combinations with distinct $xy$. It should be divided by $2=frac2!1!1!$ to avoid double counting.

$36=binom31 4cdot3 $ stays for possible combinations with two of $xyz$ being equal and distinct from the third one. It should be divided by $3=frac3!1!2!$ to avoid double counting.

$24=4cdot3cdot2 $ stays for possible combinations with all three of $xyz$ being distinct. It should be divided by $6=frac3!1!1!1!$ to avoid double counting.

share|cite|improve this answer

edited Mar 16 at 11:37

answered Mar 16 at 10:45

useruser

5,81311031

$endgroup$

add a comment | 

$begingroup$

Let list all possible combinations together with their cardinality ($xyz$ mean numbers less than or equal to 4):
$$beginalign
&[x56]:&4\
&[xy6]:&4+frac122=10\
&[x55]:&4\
&[xy5]+[x5y]:&4+frac122=10\
&[xyz]:& 4+frac363+frac246=20\
& textall:&48
endalign
$$

Explanation:

$4$ stays for possible combinations with $xy$ or $xyz$ being equal.

$12=4cdot3$ stays for possible combinations with distinct $xy$. It should be divided by $2=frac2!1!1!$ to avoid double counting.

$36=binom31 4cdot3 $ stays for possible combinations with two of $xyz$ being equal and distinct from the third one. It should be divided by $3=frac3!1!2!$ to avoid double counting.

$24=4cdot3cdot2 $ stays for possible combinations with all three of $xyz$ being distinct. It should be divided by $6=frac3!1!1!1!$ to avoid double counting.

share|cite|improve this answer

edited Mar 16 at 11:37

answered Mar 16 at 10:45

useruser

5,81311031

$endgroup$

share|cite|improve this answer

edited Mar 16 at 11:37

answered Mar 16 at 10:45

useruser

5,81311031

answered Mar 16 at 10:45

useruser

5,81311031

answered Mar 16 at 10:45

useruser

5,81311031