How many integer solutions are there to the equation $x_1+x_2+x_3+2x_4+x_5=72$?
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If the question given is to find the number of integer solutions to the equation $x_1+x_2+x_3+x_4+x_5=72$ where $x_1ge2, x_2,x_3ge1, x_4,x_5ge0$
I know that the solution would be:
$(x_1-2)+(x_2-1)+(x_3-1)+(x_4)+(x_5)=72$
So, $x_1+x_2+x_3+x_4+x_5=76$
And the number of integer solutions would be ${76+5-1 choose 76}$
But how would I find the answer if the equation given is $x_1+x_2+x_3+2x_4+x_5=72$ with same restrictions on $x_i$?
combinatorics discrete-mathematics
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up vote
-1
down vote
favorite
If the question given is to find the number of integer solutions to the equation $x_1+x_2+x_3+x_4+x_5=72$ where $x_1ge2, x_2,x_3ge1, x_4,x_5ge0$
I know that the solution would be:
$(x_1-2)+(x_2-1)+(x_3-1)+(x_4)+(x_5)=72$
So, $x_1+x_2+x_3+x_4+x_5=76$
And the number of integer solutions would be ${76+5-1 choose 76}$
But how would I find the answer if the equation given is $x_1+x_2+x_3+2x_4+x_5=72$ with same restrictions on $x_i$?
combinatorics discrete-mathematics
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
If the question given is to find the number of integer solutions to the equation $x_1+x_2+x_3+x_4+x_5=72$ where $x_1ge2, x_2,x_3ge1, x_4,x_5ge0$
I know that the solution would be:
$(x_1-2)+(x_2-1)+(x_3-1)+(x_4)+(x_5)=72$
So, $x_1+x_2+x_3+x_4+x_5=76$
And the number of integer solutions would be ${76+5-1 choose 76}$
But how would I find the answer if the equation given is $x_1+x_2+x_3+2x_4+x_5=72$ with same restrictions on $x_i$?
combinatorics discrete-mathematics
If the question given is to find the number of integer solutions to the equation $x_1+x_2+x_3+x_4+x_5=72$ where $x_1ge2, x_2,x_3ge1, x_4,x_5ge0$
I know that the solution would be:
$(x_1-2)+(x_2-1)+(x_3-1)+(x_4)+(x_5)=72$
So, $x_1+x_2+x_3+x_4+x_5=76$
And the number of integer solutions would be ${76+5-1 choose 76}$
But how would I find the answer if the equation given is $x_1+x_2+x_3+2x_4+x_5=72$ with same restrictions on $x_i$?
combinatorics discrete-mathematics
combinatorics discrete-mathematics
asked Dec 3 at 4:04
user394222
16
16
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1 Answer
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One simple approach would be to assume $x_4=k$ and solve the problem for $x_1+x_2+x_3+x_5=72-2k$ and sum over all $k$.
A more general approach would be to construct ordinary generating functions for each variable in question and find the coefficient of $x^{72}$ in their common generating function.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
One simple approach would be to assume $x_4=k$ and solve the problem for $x_1+x_2+x_3+x_5=72-2k$ and sum over all $k$.
A more general approach would be to construct ordinary generating functions for each variable in question and find the coefficient of $x^{72}$ in their common generating function.
add a comment |
up vote
1
down vote
One simple approach would be to assume $x_4=k$ and solve the problem for $x_1+x_2+x_3+x_5=72-2k$ and sum over all $k$.
A more general approach would be to construct ordinary generating functions for each variable in question and find the coefficient of $x^{72}$ in their common generating function.
add a comment |
up vote
1
down vote
up vote
1
down vote
One simple approach would be to assume $x_4=k$ and solve the problem for $x_1+x_2+x_3+x_5=72-2k$ and sum over all $k$.
A more general approach would be to construct ordinary generating functions for each variable in question and find the coefficient of $x^{72}$ in their common generating function.
One simple approach would be to assume $x_4=k$ and solve the problem for $x_1+x_2+x_3+x_5=72-2k$ and sum over all $k$.
A more general approach would be to construct ordinary generating functions for each variable in question and find the coefficient of $x^{72}$ in their common generating function.
answered Dec 3 at 4:11
gt6989b
32.6k22351
32.6k22351
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