Mixing
Possible Pool Table Layouts
$begingroup$
There is a pool game called Nine-ball. It involves 9 numbered balls 1-9 and a cue ball, so 10 balls overall. A player starts the game with a "break" shot by hitting the cue ball into the "rack of 9 balls. The question is, how many possible layouts of balls are there after the break shot?
Assumptions:
Assume the table has a resolution of one ball; that is the table is x balls wide and y balls long. A standard "9 foot" table is about 24 balls wide and 48 balls long. So there are a total of 1152 possible positions for the balls on the table (and 1 position "off the table", regardless of which pocket a ball goes in, or if it went sailing off the table on to the floor!). Obviously no two balls can occupy the same space.
There are 10 distinct balls.
Any number of balls may be pocketed on the break shot. So the maximum number of balls on the table is 10 and the minimum is zero.
How many possible layouts of the table are there immediately after the break?
combinatorics
asked Sep 14 '14 at 19:05
Pool SharkPool Shark
91
$endgroup$
add a comment |
$begingroup$
There is a pool game called Nine-ball. It involves 9 numbered balls 1-9 and a cue ball, so 10 balls overall. A player starts the game with a "break" shot by hitting the cue ball into the "rack of 9 balls. The question is, how many possible layouts of balls are there after the break shot?
Assumptions:
Assume the table has a resolution of one ball; that is the table is x balls wide and y balls long. A standard "9 foot" table is about 24 balls wide and 48 balls long. So there are a total of 1152 possible positions for the balls on the table (and 1 position "off the table", regardless of which pocket a ball goes in, or if it went sailing off the table on to the floor!). Obviously no two balls can occupy the same space.
There are 10 distinct balls.
Any number of balls may be pocketed on the break shot. So the maximum number of balls on the table is 10 and the minimum is zero.
How many possible layouts of the table are there immediately after the break?
combinatorics
asked Sep 14 '14 at 19:05
Pool SharkPool Shark
91
$endgroup$
add a comment |
$begingroup$
There is a pool game called Nine-ball. It involves 9 numbered balls 1-9 and a cue ball, so 10 balls overall. A player starts the game with a "break" shot by hitting the cue ball into the "rack of 9 balls. The question is, how many possible layouts of balls are there after the break shot?
Assumptions:
Assume the table has a resolution of one ball; that is the table is x balls wide and y balls long. A standard "9 foot" table is about 24 balls wide and 48 balls long. So there are a total of 1152 possible positions for the balls on the table (and 1 position "off the table", regardless of which pocket a ball goes in, or if it went sailing off the table on to the floor!). Obviously no two balls can occupy the same space.
There are 10 distinct balls.
Any number of balls may be pocketed on the break shot. So the maximum number of balls on the table is 10 and the minimum is zero.
How many possible layouts of the table are there immediately after the break?
combinatorics
asked Sep 14 '14 at 19:05
Pool SharkPool Shark
91
$endgroup$
There is a pool game called Nine-ball. It involves 9 numbered balls 1-9 and a cue ball, so 10 balls overall. A player starts the game with a "break" shot by hitting the cue ball into the "rack of 9 balls. The question is, how many possible layouts of balls are there after the break shot?
Assumptions:
Assume the table has a resolution of one ball; that is the table is x balls wide and y balls long. A standard "9 foot" table is about 24 balls wide and 48 balls long. So there are a total of 1152 possible positions for the balls on the table (and 1 position "off the table", regardless of which pocket a ball goes in, or if it went sailing off the table on to the floor!). Obviously no two balls can occupy the same space.
There are 10 distinct balls.
Any number of balls may be pocketed on the break shot. So the maximum number of balls on the table is 10 and the minimum is zero.
How many possible layouts of the table are there immediately after the break?
combinatorics
combinatorics
asked Sep 14 '14 at 19:05
Pool SharkPool Shark
91
asked Sep 14 '14 at 19:05
Pool SharkPool Shark
91
asked Sep 14 '14 at 19:05
Pool SharkPool Shark
91
asked Sep 14 '14 at 19:05
Pool SharkPool Shark
91
asked Sep 14 '14 at 19:05
Pool SharkPool Shark
91
91
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Lets just split into diffrent scenarios depend on how many balls were pocketed.
let $k$ be that number, then obviously we have $10-k$ balls left, in each pick of those balls there's a different table, so we first pick those balls- $10choose k$
Now, we have $10-k$ balls to arrange, lets just pick the places on the table that they occupied, we have $48cdot 24choose 10-k$ to pick them.
and last, we need to arrange those balls in the places we picked, so we have $(10-k)!$ to do that (we permutate them in a row and the left most is set to the "lowest" spot- by lex order and etc.).
So sum it up we have:
$$sum_{k=0}^{10} {10choose k}{48cdot 24choose 10-k}(10-k)!$$
answered Sep 14 '14 at 20:16
SnufsanSnufsan
1,841618
$endgroup$
$begingroup$
So how many is that? I can't do math, but it seems to me to be a variable answer; variable depending an on the number of balls pocketed, but that part of the possibilities I want to know. I don't want to know how many possibilities there are with 10 balls, or 7 balls, I want to know how many total possible layouts there are with any amount of balls (0 - 10), a number, not a formula.
$endgroup$
– Pool Shark
Sep 14 '14 at 21:27
$begingroup$
It's not a variable, just open up the sum to 11 summands and substitute k with its appropriate value for each summand
$endgroup$
– Snufsan
Sep 14 '14 at 21:48
$begingroup$
I have no idea what you just said. The forumula you provided makes no sense to me whatsoever. I am a math dunce. Can you just give me a number?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:06
$begingroup$
No! Lets say that $$f(k)={10choose k}{48cdot 24choose 10-k}(10-k)!$$ then your number is $$f(0)+f(1)+...+f(10)$$
$endgroup$
– Snufsan
Sep 14 '14 at 22:11
$begingroup$
In this formula, is that 10 divided by k? 1152 divided by 10 minus k?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:19
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Lets just split into diffrent scenarios depend on how many balls were pocketed.
let $k$ be that number, then obviously we have $10-k$ balls left, in each pick of those balls there's a different table, so we first pick those balls- $10choose k$
Now, we have $10-k$ balls to arrange, lets just pick the places on the table that they occupied, we have $48cdot 24choose 10-k$ to pick them.
and last, we need to arrange those balls in the places we picked, so we have $(10-k)!$ to do that (we permutate them in a row and the left most is set to the "lowest" spot- by lex order and etc.).
So sum it up we have:
$$sum_{k=0}^{10} {10choose k}{48cdot 24choose 10-k}(10-k)!$$
answered Sep 14 '14 at 20:16
SnufsanSnufsan
1,841618
$endgroup$
$begingroup$
So how many is that? I can't do math, but it seems to me to be a variable answer; variable depending an on the number of balls pocketed, but that part of the possibilities I want to know. I don't want to know how many possibilities there are with 10 balls, or 7 balls, I want to know how many total possible layouts there are with any amount of balls (0 - 10), a number, not a formula.
$endgroup$
– Pool Shark
Sep 14 '14 at 21:27
$begingroup$
It's not a variable, just open up the sum to 11 summands and substitute k with its appropriate value for each summand
$endgroup$
– Snufsan
Sep 14 '14 at 21:48
$begingroup$
I have no idea what you just said. The forumula you provided makes no sense to me whatsoever. I am a math dunce. Can you just give me a number?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:06
$begingroup$
No! Lets say that $$f(k)={10choose k}{48cdot 24choose 10-k}(10-k)!$$ then your number is $$f(0)+f(1)+...+f(10)$$
$endgroup$
– Snufsan
Sep 14 '14 at 22:11
$begingroup$
In this formula, is that 10 divided by k? 1152 divided by 10 minus k?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:19
|
show 2 more comments
$begingroup$
Lets just split into diffrent scenarios depend on how many balls were pocketed.
let $k$ be that number, then obviously we have $10-k$ balls left, in each pick of those balls there's a different table, so we first pick those balls- $10choose k$
Now, we have $10-k$ balls to arrange, lets just pick the places on the table that they occupied, we have $48cdot 24choose 10-k$ to pick them.
and last, we need to arrange those balls in the places we picked, so we have $(10-k)!$ to do that (we permutate them in a row and the left most is set to the "lowest" spot- by lex order and etc.).
So sum it up we have:
$$sum_{k=0}^{10} {10choose k}{48cdot 24choose 10-k}(10-k)!$$
answered Sep 14 '14 at 20:16
SnufsanSnufsan
1,841618
$endgroup$
$begingroup$
So how many is that? I can't do math, but it seems to me to be a variable answer; variable depending an on the number of balls pocketed, but that part of the possibilities I want to know. I don't want to know how many possibilities there are with 10 balls, or 7 balls, I want to know how many total possible layouts there are with any amount of balls (0 - 10), a number, not a formula.
$endgroup$
– Pool Shark
Sep 14 '14 at 21:27
$begingroup$
It's not a variable, just open up the sum to 11 summands and substitute k with its appropriate value for each summand
$endgroup$
– Snufsan
Sep 14 '14 at 21:48
$begingroup$
I have no idea what you just said. The forumula you provided makes no sense to me whatsoever. I am a math dunce. Can you just give me a number?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:06
$begingroup$
No! Lets say that $$f(k)={10choose k}{48cdot 24choose 10-k}(10-k)!$$ then your number is $$f(0)+f(1)+...+f(10)$$
$endgroup$
– Snufsan
Sep 14 '14 at 22:11
$begingroup$
In this formula, is that 10 divided by k? 1152 divided by 10 minus k?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:19
|
show 2 more comments
$begingroup$
Lets just split into diffrent scenarios depend on how many balls were pocketed.
let $k$ be that number, then obviously we have $10-k$ balls left, in each pick of those balls there's a different table, so we first pick those balls- $10choose k$
Now, we have $10-k$ balls to arrange, lets just pick the places on the table that they occupied, we have $48cdot 24choose 10-k$ to pick them.
and last, we need to arrange those balls in the places we picked, so we have $(10-k)!$ to do that (we permutate them in a row and the left most is set to the "lowest" spot- by lex order and etc.).
So sum it up we have:
$$sum_{k=0}^{10} {10choose k}{48cdot 24choose 10-k}(10-k)!$$
answered Sep 14 '14 at 20:16
SnufsanSnufsan
1,841618
$endgroup$
Lets just split into diffrent scenarios depend on how many balls were pocketed.
let $k$ be that number, then obviously we have $10-k$ balls left, in each pick of those balls there's a different table, so we first pick those balls- $10choose k$
Now, we have $10-k$ balls to arrange, lets just pick the places on the table that they occupied, we have $48cdot 24choose 10-k$ to pick them.
and last, we need to arrange those balls in the places we picked, so we have $(10-k)!$ to do that (we permutate them in a row and the left most is set to the "lowest" spot- by lex order and etc.).
So sum it up we have:
$$sum_{k=0}^{10} {10choose k}{48cdot 24choose 10-k}(10-k)!$$
answered Sep 14 '14 at 20:16
SnufsanSnufsan
1,841618
answered Sep 14 '14 at 20:16
SnufsanSnufsan
1,841618
answered Sep 14 '14 at 20:16
SnufsanSnufsan
1,841618
answered Sep 14 '14 at 20:16
SnufsanSnufsan
1,841618
1,841618
$begingroup$
So how many is that? I can't do math, but it seems to me to be a variable answer; variable depending an on the number of balls pocketed, but that part of the possibilities I want to know. I don't want to know how many possibilities there are with 10 balls, or 7 balls, I want to know how many total possible layouts there are with any amount of balls (0 - 10), a number, not a formula.
$endgroup$
– Pool Shark
Sep 14 '14 at 21:27
$begingroup$
It's not a variable, just open up the sum to 11 summands and substitute k with its appropriate value for each summand
$endgroup$
– Snufsan
Sep 14 '14 at 21:48
$begingroup$
I have no idea what you just said. The forumula you provided makes no sense to me whatsoever. I am a math dunce. Can you just give me a number?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:06
$begingroup$
No! Lets say that $$f(k)={10choose k}{48cdot 24choose 10-k}(10-k)!$$ then your number is $$f(0)+f(1)+...+f(10)$$
$endgroup$
– Snufsan
Sep 14 '14 at 22:11
$begingroup$
In this formula, is that 10 divided by k? 1152 divided by 10 minus k?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:19
|
show 2 more comments
$begingroup$
So how many is that? I can't do math, but it seems to me to be a variable answer; variable depending an on the number of balls pocketed, but that part of the possibilities I want to know. I don't want to know how many possibilities there are with 10 balls, or 7 balls, I want to know how many total possible layouts there are with any amount of balls (0 - 10), a number, not a formula.
$endgroup$
– Pool Shark
Sep 14 '14 at 21:27
$begingroup$
It's not a variable, just open up the sum to 11 summands and substitute k with its appropriate value for each summand
$endgroup$
– Snufsan
Sep 14 '14 at 21:48
$begingroup$
I have no idea what you just said. The forumula you provided makes no sense to me whatsoever. I am a math dunce. Can you just give me a number?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:06
$begingroup$
No! Lets say that $$f(k)={10choose k}{48cdot 24choose 10-k}(10-k)!$$ then your number is $$f(0)+f(1)+...+f(10)$$
$endgroup$
– Snufsan
Sep 14 '14 at 22:11
$begingroup$
In this formula, is that 10 divided by k? 1152 divided by 10 minus k?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:19
$begingroup$
So how many is that? I can't do math, but it seems to me to be a variable answer; variable depending an on the number of balls pocketed, but that part of the possibilities I want to know. I don't want to know how many possibilities there are with 10 balls, or 7 balls, I want to know how many total possible layouts there are with any amount of balls (0 - 10), a number, not a formula.
$endgroup$
– Pool Shark
Sep 14 '14 at 21:27
$begingroup$
So how many is that? I can't do math, but it seems to me to be a variable answer; variable depending an on the number of balls pocketed, but that part of the possibilities I want to know. I don't want to know how many possibilities there are with 10 balls, or 7 balls, I want to know how many total possible layouts there are with any amount of balls (0 - 10), a number, not a formula.
$endgroup$
– Pool Shark
Sep 14 '14 at 21:27
$begingroup$
It's not a variable, just open up the sum to 11 summands and substitute k with its appropriate value for each summand
$endgroup$
– Snufsan
Sep 14 '14 at 21:48
$begingroup$
It's not a variable, just open up the sum to 11 summands and substitute k with its appropriate value for each summand
$endgroup$
– Snufsan
Sep 14 '14 at 21:48
$begingroup$
I have no idea what you just said. The forumula you provided makes no sense to me whatsoever. I am a math dunce. Can you just give me a number?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:06
$begingroup$
I have no idea what you just said. The forumula you provided makes no sense to me whatsoever. I am a math dunce. Can you just give me a number?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:06
$begingroup$
No! Lets say that $$f(k)={10choose k}{48cdot 24choose 10-k}(10-k)!$$ then your number is $$f(0)+f(1)+...+f(10)$$
$endgroup$
– Snufsan
Sep 14 '14 at 22:11
$begingroup$
No! Lets say that $$f(k)={10choose k}{48cdot 24choose 10-k}(10-k)!$$ then your number is $$f(0)+f(1)+...+f(10)$$
$endgroup$
– Snufsan
Sep 14 '14 at 22:11
$begingroup$
In this formula, is that 10 divided by k? 1152 divided by 10 minus k?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:19
$begingroup$
In this formula, is that 10 divided by k? 1152 divided by 10 minus k?
$endgroup$
– Pool Shark
Sep 14 '14 at 22:19
|
show 2 more comments
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