Mixing
number of solutions for this equation
$begingroup$
What is the number of solutions for this equation:
$$x_1 + x_2 + x_3 = 15$$
when $x_1 in mathbb{N}_{even} cup {0}$, $0leq x_2leq 5$, $x_3 geq 0$?
Can it be solved without generating functions? If not, how can I use generating functions to solve it?
EDIT: $x_1, x_2, x_3$ can be only integers
combinatorics combinations
asked Jan 12 at 17:50
Robo YonuomaroRobo Yonuomaro
786
$endgroup$
add a comment |
$begingroup$
What is the number of solutions for this equation:
$$x_1 + x_2 + x_3 = 15$$
when $x_1 in mathbb{N}_{even} cup {0}$, $0leq x_2leq 5$, $x_3 geq 0$?
Can it be solved without generating functions? If not, how can I use generating functions to solve it?
EDIT: $x_1, x_2, x_3$ can be only integers
combinatorics combinations
asked Jan 12 at 17:50
Robo YonuomaroRobo Yonuomaro
786
$endgroup$
2
$begingroup$
There are so few possibilities you can just list them.
$endgroup$
– saulspatz
Jan 12 at 17:53
$begingroup$
@saulspatz I have counted at least 20, it seems like a lot of work
$endgroup$
– Robo Yonuomaro
Jan 12 at 17:58
$begingroup$
Can $x_2,x_3$ be real or just integers are alowed?
$endgroup$
– Patricio
Jan 12 at 18:09
$begingroup$
Integers, I forgot to mention it
$endgroup$
– Robo Yonuomaro
Jan 12 at 18:10
add a comment |
$begingroup$
What is the number of solutions for this equation:
$$x_1 + x_2 + x_3 = 15$$
when $x_1 in mathbb{N}_{even} cup {0}$, $0leq x_2leq 5$, $x_3 geq 0$?
Can it be solved without generating functions? If not, how can I use generating functions to solve it?
EDIT: $x_1, x_2, x_3$ can be only integers
combinatorics combinations
asked Jan 12 at 17:50
Robo YonuomaroRobo Yonuomaro
786
$endgroup$
What is the number of solutions for this equation:
$$x_1 + x_2 + x_3 = 15$$
when $x_1 in mathbb{N}_{even} cup {0}$, $0leq x_2leq 5$, $x_3 geq 0$?
Can it be solved without generating functions? If not, how can I use generating functions to solve it?
EDIT: $x_1, x_2, x_3$ can be only integers
combinatorics combinations
combinatorics combinations
asked Jan 12 at 17:50
Robo YonuomaroRobo Yonuomaro
786
asked Jan 12 at 17:50
Robo YonuomaroRobo Yonuomaro
786
edited Jan 12 at 18:11
Robo Yonuomaro
asked Jan 12 at 17:50
Robo YonuomaroRobo Yonuomaro
786
asked Jan 12 at 17:50
Robo YonuomaroRobo Yonuomaro
786
asked Jan 12 at 17:50
Robo YonuomaroRobo Yonuomaro
786
786
2
$begingroup$
There are so few possibilities you can just list them.
$endgroup$
– saulspatz
Jan 12 at 17:53
$begingroup$
@saulspatz I have counted at least 20, it seems like a lot of work
$endgroup$
– Robo Yonuomaro
Jan 12 at 17:58
$begingroup$
Can $x_2,x_3$ be real or just integers are alowed?
$endgroup$
– Patricio
Jan 12 at 18:09
$begingroup$
Integers, I forgot to mention it
$endgroup$
– Robo Yonuomaro
Jan 12 at 18:10
add a comment |
2
$begingroup$
There are so few possibilities you can just list them.
$endgroup$
– saulspatz
Jan 12 at 17:53
$begingroup$
@saulspatz I have counted at least 20, it seems like a lot of work
$endgroup$
– Robo Yonuomaro
Jan 12 at 17:58
$begingroup$
Can $x_2,x_3$ be real or just integers are alowed?
$endgroup$
– Patricio
Jan 12 at 18:09
$begingroup$
Integers, I forgot to mention it
$endgroup$
– Robo Yonuomaro
Jan 12 at 18:10
2
2
$begingroup$
There are so few possibilities you can just list them.
$endgroup$
– saulspatz
Jan 12 at 17:53
$begingroup$
There are so few possibilities you can just list them.
$endgroup$
– saulspatz
Jan 12 at 17:53
$begingroup$
@saulspatz I have counted at least 20, it seems like a lot of work
$endgroup$
– Robo Yonuomaro
Jan 12 at 17:58
$begingroup$
@saulspatz I have counted at least 20, it seems like a lot of work
$endgroup$
– Robo Yonuomaro
Jan 12 at 17:58
$begingroup$
Can $x_2,x_3$ be real or just integers are alowed?
$endgroup$
– Patricio
Jan 12 at 18:09
$begingroup$
Can $x_2,x_3$ be real or just integers are alowed?
$endgroup$
– Patricio
Jan 12 at 18:09
$begingroup$
Integers, I forgot to mention it
$endgroup$
– Robo Yonuomaro
Jan 12 at 18:10
$begingroup$
Integers, I forgot to mention it
$endgroup$
– Robo Yonuomaro
Jan 12 at 18:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Start by putting $x_1=2k$ (where $k$ varies from $0$ to $7$). This gives us the following
$$x_2 + x_3 = 15-2k = n$$
Now, using the stars and bars rule we know that the above equation has following number of integral solutions for any particular $k$
$$N_1 = binom{n+2-1}{2-1} = binom{n+1}{1} = binom{16+2k}{1}$$
Now from this we subtract all the cases in which $x_2ge6$. So put $x_2=y_2+6$ which gives us $y_2ge0$
$$y_2 + x_3 = 15-2k-6 = 9-2k$$
Number of integral solutions to this are
$$N_2 = binom{9-2k+2-1}{2-1} = binom{10-2k}{1}$$
Summing up $N_1$ from $k=0$ to $k=7$
$$sum_{k=0}^7binom{16+2k}{1} =sum_{k=0}^716+2k=184$$
Now summing up $N_2$ from $k=0$ to $k=4$ as after that the terms will simply be $0$ in the binomial coefficient
$$sum_{k=0}^4binom{10-2k}{1} =sum_{k=0}^410-2k = 30$$
Subtracting $N_2$ from $N_1$, we'd get
$$N=N_1-N_2 = 184-30 =154$$
which is the total number of possibilities.
answered Jan 13 at 8:14
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3071188%2fnumber-of-solutions-for-this-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Start by putting $x_1=2k$ (where $k$ varies from $0$ to $7$). This gives us the following
$$x_2 + x_3 = 15-2k = n$$
Now, using the stars and bars rule we know that the above equation has following number of integral solutions for any particular $k$
$$N_1 = binom{n+2-1}{2-1} = binom{n+1}{1} = binom{16+2k}{1}$$
Now from this we subtract all the cases in which $x_2ge6$. So put $x_2=y_2+6$ which gives us $y_2ge0$
$$y_2 + x_3 = 15-2k-6 = 9-2k$$
Number of integral solutions to this are
$$N_2 = binom{9-2k+2-1}{2-1} = binom{10-2k}{1}$$
Summing up $N_1$ from $k=0$ to $k=7$
$$sum_{k=0}^7binom{16+2k}{1} =sum_{k=0}^716+2k=184$$
Now summing up $N_2$ from $k=0$ to $k=4$ as after that the terms will simply be $0$ in the binomial coefficient
$$sum_{k=0}^4binom{10-2k}{1} =sum_{k=0}^410-2k = 30$$
Subtracting $N_2$ from $N_1$, we'd get
$$N=N_1-N_2 = 184-30 =154$$
which is the total number of possibilities.
answered Jan 13 at 8:14
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
Start by putting $x_1=2k$ (where $k$ varies from $0$ to $7$). This gives us the following
$$x_2 + x_3 = 15-2k = n$$
Now, using the stars and bars rule we know that the above equation has following number of integral solutions for any particular $k$
$$N_1 = binom{n+2-1}{2-1} = binom{n+1}{1} = binom{16+2k}{1}$$
Now from this we subtract all the cases in which $x_2ge6$. So put $x_2=y_2+6$ which gives us $y_2ge0$
$$y_2 + x_3 = 15-2k-6 = 9-2k$$
Number of integral solutions to this are
$$N_2 = binom{9-2k+2-1}{2-1} = binom{10-2k}{1}$$
Summing up $N_1$ from $k=0$ to $k=7$
$$sum_{k=0}^7binom{16+2k}{1} =sum_{k=0}^716+2k=184$$
Now summing up $N_2$ from $k=0$ to $k=4$ as after that the terms will simply be $0$ in the binomial coefficient
$$sum_{k=0}^4binom{10-2k}{1} =sum_{k=0}^410-2k = 30$$
Subtracting $N_2$ from $N_1$, we'd get
$$N=N_1-N_2 = 184-30 =154$$
which is the total number of possibilities.
answered Jan 13 at 8:14
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
Start by putting $x_1=2k$ (where $k$ varies from $0$ to $7$). This gives us the following
$$x_2 + x_3 = 15-2k = n$$
Now, using the stars and bars rule we know that the above equation has following number of integral solutions for any particular $k$
$$N_1 = binom{n+2-1}{2-1} = binom{n+1}{1} = binom{16+2k}{1}$$
Now from this we subtract all the cases in which $x_2ge6$. So put $x_2=y_2+6$ which gives us $y_2ge0$
$$y_2 + x_3 = 15-2k-6 = 9-2k$$
Number of integral solutions to this are
$$N_2 = binom{9-2k+2-1}{2-1} = binom{10-2k}{1}$$
Summing up $N_1$ from $k=0$ to $k=7$
$$sum_{k=0}^7binom{16+2k}{1} =sum_{k=0}^716+2k=184$$
Now summing up $N_2$ from $k=0$ to $k=4$ as after that the terms will simply be $0$ in the binomial coefficient
$$sum_{k=0}^4binom{10-2k}{1} =sum_{k=0}^410-2k = 30$$
Subtracting $N_2$ from $N_1$, we'd get
$$N=N_1-N_2 = 184-30 =154$$
which is the total number of possibilities.
answered Jan 13 at 8:14
Sauhard SharmaSauhard Sharma
953318
$endgroup$
Start by putting $x_1=2k$ (where $k$ varies from $0$ to $7$). This gives us the following
$$x_2 + x_3 = 15-2k = n$$
Now, using the stars and bars rule we know that the above equation has following number of integral solutions for any particular $k$
$$N_1 = binom{n+2-1}{2-1} = binom{n+1}{1} = binom{16+2k}{1}$$
Now from this we subtract all the cases in which $x_2ge6$. So put $x_2=y_2+6$ which gives us $y_2ge0$
$$y_2 + x_3 = 15-2k-6 = 9-2k$$
Number of integral solutions to this are
$$N_2 = binom{9-2k+2-1}{2-1} = binom{10-2k}{1}$$
Summing up $N_1$ from $k=0$ to $k=7$
$$sum_{k=0}^7binom{16+2k}{1} =sum_{k=0}^716+2k=184$$
Now summing up $N_2$ from $k=0$ to $k=4$ as after that the terms will simply be $0$ in the binomial coefficient
$$sum_{k=0}^4binom{10-2k}{1} =sum_{k=0}^410-2k = 30$$
Subtracting $N_2$ from $N_1$, we'd get
$$N=N_1-N_2 = 184-30 =154$$
which is the total number of possibilities.
answered Jan 13 at 8:14
Sauhard SharmaSauhard Sharma
953318
edited Jan 14 at 8:35
answered Jan 13 at 8:14
Sauhard SharmaSauhard Sharma
953318
answered Jan 13 at 8:14
Sauhard SharmaSauhard Sharma
953318
answered Jan 13 at 8:14
Sauhard SharmaSauhard Sharma
953318
953318
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3071188%2fnumber-of-solutions-for-this-equation%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
2
$begingroup$
There are so few possibilities you can just list them.
$endgroup$
– saulspatz
Jan 12 at 17:53
$begingroup$
@saulspatz I have counted at least 20, it seems like a lot of work
$endgroup$
– Robo Yonuomaro
Jan 12 at 17:58
$begingroup$
Can $x_2,x_3$ be real or just integers are alowed?
$endgroup$
– Patricio
Jan 12 at 18:09
$begingroup$
Integers, I forgot to mention it
$endgroup$
– Robo Yonuomaro
Jan 12 at 18:10