How many ways are there to distribute 25 identical balls among 5 players where each player must get at least...
$begingroup$
We know that
$$x_1 + x_2 + x_3 + x_4 + x_5 = 25; 1leq x_i < 10$$.
Therefore, I'm thinking about getting all possible combinations and subtracting them by where 4 people get at least 1 and one other gets at least 11 so we'd have
$$x_1 + x_2 + x_3 + x_4 + x_5 = 11$$
Getting all the possible combinations (multiplying by 5 since any one of them could have more than 10 balls)
And subtract both of the answers.
combinatorics discrete-mathematics
edited Dec 10 '18 at 21:24
mm-crj
425213
asked Dec 10 '18 at 21:09
NinjaWarrriorNinjaWarrrior
62
$endgroup$
add a comment |
$begingroup$
We know that
$$x_1 + x_2 + x_3 + x_4 + x_5 = 25; 1leq x_i < 10$$.
Therefore, I'm thinking about getting all possible combinations and subtracting them by where 4 people get at least 1 and one other gets at least 11 so we'd have
$$x_1 + x_2 + x_3 + x_4 + x_5 = 11$$
Getting all the possible combinations (multiplying by 5 since any one of them could have more than 10 balls)
And subtract both of the answers.
combinatorics discrete-mathematics
edited Dec 10 '18 at 21:24
mm-crj
425213
asked Dec 10 '18 at 21:09
NinjaWarrriorNinjaWarrrior
62
$endgroup$
1
$begingroup$
Please show us what you have attempted. Since two people could simultaneously receive $10$ or more balls, subtracting the number of distributions in which a single person receives $10$ or more balls results in subtracting those arrangements in which two people receive ten or more balls twice.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:59
$begingroup$
@N. F. Taussig I should not have given him the fish so fast .. no? :D
$endgroup$
– mm-crj
Dec 10 '18 at 22:15
$begingroup$
Have a look at the generating functions method, other examples of which there are plenty on MSE
$endgroup$
– rtybase
Dec 10 '18 at 22:20
add a comment |
$begingroup$
We know that
$$x_1 + x_2 + x_3 + x_4 + x_5 = 25; 1leq x_i < 10$$.
Therefore, I'm thinking about getting all possible combinations and subtracting them by where 4 people get at least 1 and one other gets at least 11 so we'd have
$$x_1 + x_2 + x_3 + x_4 + x_5 = 11$$
Getting all the possible combinations (multiplying by 5 since any one of them could have more than 10 balls)
And subtract both of the answers.
combinatorics discrete-mathematics
edited Dec 10 '18 at 21:24
mm-crj
425213
asked Dec 10 '18 at 21:09
NinjaWarrriorNinjaWarrrior
62
$endgroup$
We know that
$$x_1 + x_2 + x_3 + x_4 + x_5 = 25; 1leq x_i < 10$$.
Therefore, I'm thinking about getting all possible combinations and subtracting them by where 4 people get at least 1 and one other gets at least 11 so we'd have
$$x_1 + x_2 + x_3 + x_4 + x_5 = 11$$
Getting all the possible combinations (multiplying by 5 since any one of them could have more than 10 balls)
And subtract both of the answers.
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Dec 10 '18 at 21:24
mm-crj
425213
asked Dec 10 '18 at 21:09
NinjaWarrriorNinjaWarrrior
62
edited Dec 10 '18 at 21:24
mm-crj
425213
asked Dec 10 '18 at 21:09
NinjaWarrriorNinjaWarrrior
62
edited Dec 10 '18 at 21:24
mm-crj
425213
edited Dec 10 '18 at 21:24
mm-crj
425213
edited Dec 10 '18 at 21:24
mm-crj
425213
425213
asked Dec 10 '18 at 21:09
NinjaWarrriorNinjaWarrrior
62
asked Dec 10 '18 at 21:09
NinjaWarrriorNinjaWarrrior
62
asked Dec 10 '18 at 21:09
NinjaWarrriorNinjaWarrrior
62
62
1
$begingroup$
Please show us what you have attempted. Since two people could simultaneously receive $10$ or more balls, subtracting the number of distributions in which a single person receives $10$ or more balls results in subtracting those arrangements in which two people receive ten or more balls twice.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:59
$begingroup$
@N. F. Taussig I should not have given him the fish so fast .. no? :D
$endgroup$
– mm-crj
Dec 10 '18 at 22:15
$begingroup$
Have a look at the generating functions method, other examples of which there are plenty on MSE
$endgroup$
– rtybase
Dec 10 '18 at 22:20
add a comment |
1
$begingroup$
Please show us what you have attempted. Since two people could simultaneously receive $10$ or more balls, subtracting the number of distributions in which a single person receives $10$ or more balls results in subtracting those arrangements in which two people receive ten or more balls twice.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:59
$begingroup$
@N. F. Taussig I should not have given him the fish so fast .. no? :D
$endgroup$
– mm-crj
Dec 10 '18 at 22:15
$begingroup$
Have a look at the generating functions method, other examples of which there are plenty on MSE
$endgroup$
– rtybase
Dec 10 '18 at 22:20
1
1
$begingroup$
Please show us what you have attempted. Since two people could simultaneously receive $10$ or more balls, subtracting the number of distributions in which a single person receives $10$ or more balls results in subtracting those arrangements in which two people receive ten or more balls twice.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:59
$begingroup$
Please show us what you have attempted. Since two people could simultaneously receive $10$ or more balls, subtracting the number of distributions in which a single person receives $10$ or more balls results in subtracting those arrangements in which two people receive ten or more balls twice.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:59
$begingroup$
@N. F. Taussig I should not have given him the fish so fast .. no? :D
$endgroup$
– mm-crj
Dec 10 '18 at 22:15
$begingroup$
@N. F. Taussig I should not have given him the fish so fast .. no? :D
$endgroup$
– mm-crj
Dec 10 '18 at 22:15
$begingroup$
Have a look at the generating functions method, other examples of which there are plenty on MSE
$endgroup$
– rtybase
Dec 10 '18 at 22:20
$begingroup$
Have a look at the generating functions method, other examples of which there are plenty on MSE
$endgroup$
– rtybase
Dec 10 '18 at 22:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An intuitive way to think about this is to consider a problem of arranging 0's and 1's.
Say you have n balls to distributed amongst k people with the same conditions as above. That each person should get atleast $1$ ball and the number of balls with one person can be at max $l$. Therefore the problem is as follows:
You have $n$ 1's and $k-1$ zeros to arrange in such a way that the zeros don't lie in the ends and zeros are not adjacent. This ensures the minimum 1 ball per person condition. The sum of the of the 1's between zeros is the number of balls for that person.
Example:
$$111011110110cdots$$
Therefore $x_1=3$, $x_2=4$, $x_3=2$ and so on. There are $n-1$ holes to be filled amongst the $n$ 1's by $k-1$ zeros. And this can be done in ${n-1}choose{k-1}$ ways.
And about each person having a maximum of $l-1$ balls condition can be enforced by subtracting the cases in which one person has $l$ or more balls. This is easy in this case because only 2 people can have 11 balls at a time$(11+11+3=25)$.
answered Dec 10 '18 at 21:37
mm-crjmm-crj
425213
$endgroup$
add a comment |
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$begingroup$
An intuitive way to think about this is to consider a problem of arranging 0's and 1's.
Say you have n balls to distributed amongst k people with the same conditions as above. That each person should get atleast $1$ ball and the number of balls with one person can be at max $l$. Therefore the problem is as follows:
You have $n$ 1's and $k-1$ zeros to arrange in such a way that the zeros don't lie in the ends and zeros are not adjacent. This ensures the minimum 1 ball per person condition. The sum of the of the 1's between zeros is the number of balls for that person.
Example:
$$111011110110cdots$$
Therefore $x_1=3$, $x_2=4$, $x_3=2$ and so on. There are $n-1$ holes to be filled amongst the $n$ 1's by $k-1$ zeros. And this can be done in ${n-1}choose{k-1}$ ways.
And about each person having a maximum of $l-1$ balls condition can be enforced by subtracting the cases in which one person has $l$ or more balls. This is easy in this case because only 2 people can have 11 balls at a time$(11+11+3=25)$.
answered Dec 10 '18 at 21:37
mm-crjmm-crj
425213
$endgroup$
add a comment |
$begingroup$
An intuitive way to think about this is to consider a problem of arranging 0's and 1's.
Say you have n balls to distributed amongst k people with the same conditions as above. That each person should get atleast $1$ ball and the number of balls with one person can be at max $l$. Therefore the problem is as follows:
You have $n$ 1's and $k-1$ zeros to arrange in such a way that the zeros don't lie in the ends and zeros are not adjacent. This ensures the minimum 1 ball per person condition. The sum of the of the 1's between zeros is the number of balls for that person.
Example:
$$111011110110cdots$$
Therefore $x_1=3$, $x_2=4$, $x_3=2$ and so on. There are $n-1$ holes to be filled amongst the $n$ 1's by $k-1$ zeros. And this can be done in ${n-1}choose{k-1}$ ways.
And about each person having a maximum of $l-1$ balls condition can be enforced by subtracting the cases in which one person has $l$ or more balls. This is easy in this case because only 2 people can have 11 balls at a time$(11+11+3=25)$.
answered Dec 10 '18 at 21:37
mm-crjmm-crj
425213
$endgroup$
add a comment |
$begingroup$
An intuitive way to think about this is to consider a problem of arranging 0's and 1's.
Say you have n balls to distributed amongst k people with the same conditions as above. That each person should get atleast $1$ ball and the number of balls with one person can be at max $l$. Therefore the problem is as follows:
You have $n$ 1's and $k-1$ zeros to arrange in such a way that the zeros don't lie in the ends and zeros are not adjacent. This ensures the minimum 1 ball per person condition. The sum of the of the 1's between zeros is the number of balls for that person.
Example:
$$111011110110cdots$$
Therefore $x_1=3$, $x_2=4$, $x_3=2$ and so on. There are $n-1$ holes to be filled amongst the $n$ 1's by $k-1$ zeros. And this can be done in ${n-1}choose{k-1}$ ways.
And about each person having a maximum of $l-1$ balls condition can be enforced by subtracting the cases in which one person has $l$ or more balls. This is easy in this case because only 2 people can have 11 balls at a time$(11+11+3=25)$.
answered Dec 10 '18 at 21:37
mm-crjmm-crj
425213
$endgroup$
An intuitive way to think about this is to consider a problem of arranging 0's and 1's.
Say you have n balls to distributed amongst k people with the same conditions as above. That each person should get atleast $1$ ball and the number of balls with one person can be at max $l$. Therefore the problem is as follows:
You have $n$ 1's and $k-1$ zeros to arrange in such a way that the zeros don't lie in the ends and zeros are not adjacent. This ensures the minimum 1 ball per person condition. The sum of the of the 1's between zeros is the number of balls for that person.
Example:
$$111011110110cdots$$
Therefore $x_1=3$, $x_2=4$, $x_3=2$ and so on. There are $n-1$ holes to be filled amongst the $n$ 1's by $k-1$ zeros. And this can be done in ${n-1}choose{k-1}$ ways.
And about each person having a maximum of $l-1$ balls condition can be enforced by subtracting the cases in which one person has $l$ or more balls. This is easy in this case because only 2 people can have 11 balls at a time$(11+11+3=25)$.
answered Dec 10 '18 at 21:37
mm-crjmm-crj
425213
answered Dec 10 '18 at 21:37
mm-crjmm-crj
425213
answered Dec 10 '18 at 21:37
mm-crjmm-crj
425213
answered Dec 10 '18 at 21:37
mm-crjmm-crj
425213
425213
add a comment |
add a comment |
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1
$begingroup$
Please show us what you have attempted. Since two people could simultaneously receive $10$ or more balls, subtracting the number of distributions in which a single person receives $10$ or more balls results in subtracting those arrangements in which two people receive ten or more balls twice.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 21:59
$begingroup$
@N. F. Taussig I should not have given him the fish so fast .. no? :D
$endgroup$
– mm-crj
Dec 10 '18 at 22:15
$begingroup$
Have a look at the generating functions method, other examples of which there are plenty on MSE
$endgroup$
– rtybase
Dec 10 '18 at 22:20