Posible Dominoes Games

Verdugo · #1 $Old$ November 5th, 2009, 03:04 PM

Hi and I would really appreciate if somebody help me on this.

I want to find out how many dominoes games are posible to do with a set of 28 tiles and every game including 4 players in which each takes 7 tiles (4x7=28

)

I was thinking the solution would be (Z):
C (28,7) x C (21,7) x C (14,7)= 4.72E+14=Z

I dont think the number is correct as if I pick a game random from Z and I call it R and I call the four players R1, R2, R3 and R4.

Lets say that our current game R made of R1, R2, R3 and R4 is changed to R4, R2, R1 and R3 (the players simply changed positions in the table, keeping their tiles). For my purposes, this is a different game, even though the players still have the same tiles but the fact that they are in a different positions, it creates a different game as the turns are anti clock.

I wonder if I have to do a permutation of 4 which is equal to 24.

Then my number of games could be equal to 24 x 4.72E+14

is this correct? or are those variations already on my previous solution?

thanks for the help.

Grandad · #2 $Old$ November 6th, 2009, 09:21 AM

Hello Verdugo

Welcome to Math Help Forum!

Quote:

Originally Posted by Verdugo $View Post$

Hi and I would really appreciate if somebody help me on this.

I want to find out how many dominoes games are posible to do with a set of 28 tiles and every game including 4 players in which each takes 7 tiles (4x7=28

)

I was thinking the solution would be (Z):
C (28,7) x C (21,7) x C (14,7)= 4.72E+14=Z

I dont think the number is correct as if I pick a game random from Z and I call it R and I call the four players R1, R2, R3 and R4.

Lets say that our current game R made of R1, R2, R3 and R4 is changed to R4, R2, R1 and R3 (the players simply changed positions in the table, keeping their tiles). For my purposes, this is a different game, even though the players still have the same tiles but the fact that they are in a different positions, it creates a different game as the turns are anti clock.

I wonder if I have to do a permutation of 4 which is equal to 24.

Then my number of games could be equal to 24 x 4.72E+14

is this correct? or are those variations already on my previous solution?

thanks for the help.

You are right. You can divide the $28$ dominoes into $4$ groups of $7$ in $\binom{28}{7}\binom{21}{7}\binom{14}{7}$ ways, and then allocate these groups to the four players in $4!$ ways. Thus the total is

$4!\binom{28}{7}\binom{21}{7}\binom{14}{7}\approx 1.134\times 10^{16}$

Grandad

Verdugo · #3 $Old$ November 6th, 2009, 11:45 AM

Quote:

Originally Posted by Grandad $View Post$

Hello Verdugo

Welcome to Math Help Forum!You are right. You can divide the $28$ dominoes into $4$ groups of $7$ in $\binom{28}{7}\binom{21}{7}\binom{14}{7}$ ways, and then allocate these groups to the four players in $4!$ ways. Thus the total is

$4!\binom{28}{7}\binom{21}{7}\binom{14}{7}\approx 1.134\times 10^{16}$

Grandad

Thank you very much for your help.