Probability that sum of three digits is the same as sum of other three digits
up vote
1
down vote
favorite
How can I solve problems like this:
Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,
$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.
I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.
probability combinatorics
add a comment |
up vote
1
down vote
favorite
How can I solve problems like this:
Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,
$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.
I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.
probability combinatorics
Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
– Keen-ameteur
Nov 18 at 14:55
1
Are the numbers chosen with or without replacement?
– Ross Millikan
Nov 18 at 14:57
Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 18 at 15:25
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How can I solve problems like this:
Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,
$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.
I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.
probability combinatorics
How can I solve problems like this:
Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,
$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.
I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.
probability combinatorics
probability combinatorics
edited Nov 18 at 16:11
N. F. Taussig
42.8k93254
42.8k93254
asked Nov 18 at 14:51
Gokuruto
123
123
Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
– Keen-ameteur
Nov 18 at 14:55
1
Are the numbers chosen with or without replacement?
– Ross Millikan
Nov 18 at 14:57
Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 18 at 15:25
add a comment |
Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
– Keen-ameteur
Nov 18 at 14:55
1
Are the numbers chosen with or without replacement?
– Ross Millikan
Nov 18 at 14:57
Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 18 at 15:25
Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
– Keen-ameteur
Nov 18 at 14:55
Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
– Keen-ameteur
Nov 18 at 14:55
1
1
Are the numbers chosen with or without replacement?
– Ross Millikan
Nov 18 at 14:57
Are the numbers chosen with or without replacement?
– Ross Millikan
Nov 18 at 14:57
Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 18 at 15:25
Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 18 at 15:25
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
The final result $p$ is then simply given by
$$p=sum_{k=0}^{27}p_k^2 .$$
By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
$$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$
add a comment |
up vote
0
down vote
You are looking for
$$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
0 leqslant text{integer }x_{,j} leqslant r hfill \
x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
end{gathered} right.$$
which is given by
$$
N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
{left( { - 1} right)^k binom{m}{k}
binom
{ s + m - 1 - kleft( {r + 1} right) }
{ s - kleft( {r + 1} right)} }
$$
as widely explained in this related post.
Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
The final result $p$ is then simply given by
$$p=sum_{k=0}^{27}p_k^2 .$$
By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
$$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$
add a comment |
up vote
2
down vote
accepted
It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
The final result $p$ is then simply given by
$$p=sum_{k=0}^{27}p_k^2 .$$
By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
$$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
The final result $p$ is then simply given by
$$p=sum_{k=0}^{27}p_k^2 .$$
By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
$$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$
It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
The final result $p$ is then simply given by
$$p=sum_{k=0}^{27}p_k^2 .$$
By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
$$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$
answered Nov 18 at 16:49
Christian Blatter
171k7111325
171k7111325
add a comment |
add a comment |
up vote
0
down vote
You are looking for
$$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
0 leqslant text{integer }x_{,j} leqslant r hfill \
x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
end{gathered} right.$$
which is given by
$$
N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
{left( { - 1} right)^k binom{m}{k}
binom
{ s + m - 1 - kleft( {r + 1} right) }
{ s - kleft( {r + 1} right)} }
$$
as widely explained in this related post.
Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.
add a comment |
up vote
0
down vote
You are looking for
$$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
0 leqslant text{integer }x_{,j} leqslant r hfill \
x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
end{gathered} right.$$
which is given by
$$
N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
{left( { - 1} right)^k binom{m}{k}
binom
{ s + m - 1 - kleft( {r + 1} right) }
{ s - kleft( {r + 1} right)} }
$$
as widely explained in this related post.
Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.
add a comment |
up vote
0
down vote
up vote
0
down vote
You are looking for
$$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
0 leqslant text{integer }x_{,j} leqslant r hfill \
x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
end{gathered} right.$$
which is given by
$$
N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
{left( { - 1} right)^k binom{m}{k}
binom
{ s + m - 1 - kleft( {r + 1} right) }
{ s - kleft( {r + 1} right)} }
$$
as widely explained in this related post.
Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.
You are looking for
$$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
0 leqslant text{integer }x_{,j} leqslant r hfill \
x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
end{gathered} right.$$
which is given by
$$
N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
{left( { - 1} right)^k binom{m}{k}
binom
{ s + m - 1 - kleft( {r + 1} right) }
{ s - kleft( {r + 1} right)} }
$$
as widely explained in this related post.
Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.
answered Nov 18 at 17:08
G Cab
17k31237
17k31237
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003625%2fprobability-that-sum-of-three-digits-is-the-same-as-sum-of-other-three-digits%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Are these supposed to be the digits of 6-digit natural number? Because if so then the first digit has to be non-zero.
– Keen-ameteur
Nov 18 at 14:55
1
Are the numbers chosen with or without replacement?
– Ross Millikan
Nov 18 at 14:57
Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering.
– N. F. Taussig
Nov 18 at 15:25