simple combinatorics: product rule and number of ordered pairs
$begingroup$
Product rule: if $X,Y$ -- finite sets and $|X|=n$, $|Y|= m$ we can say, that the number of the ways to first extract an element from the first set and then an element from the second set $= nm$ -- direct product of sets $X$ and $Y$.
${n+m}choose{2}$$2!= frac{(n+m)!}{2! (n+m-2)!} = frac{(n+m)(n+m-1)}{2}$ --the number of non-ordered pairs in set $Xcup Y$, where $|Xcup Y| = n + m$.
What the difference between these two formulas? I cannot see... Why $frac{(n+m)(n+m-1)}{2} neq nm$?
combinatorics
asked Dec 29 '18 at 0:02
Just do itJust do it
19618
$endgroup$
add a comment |
$begingroup$
Product rule: if $X,Y$ -- finite sets and $|X|=n$, $|Y|= m$ we can say, that the number of the ways to first extract an element from the first set and then an element from the second set $= nm$ -- direct product of sets $X$ and $Y$.
${n+m}choose{2}$$2!= frac{(n+m)!}{2! (n+m-2)!} = frac{(n+m)(n+m-1)}{2}$ --the number of non-ordered pairs in set $Xcup Y$, where $|Xcup Y| = n + m$.
What the difference between these two formulas? I cannot see... Why $frac{(n+m)(n+m-1)}{2} neq nm$?
combinatorics
asked Dec 29 '18 at 0:02
Just do itJust do it
19618
$endgroup$
add a comment |
$begingroup$
Product rule: if $X,Y$ -- finite sets and $|X|=n$, $|Y|= m$ we can say, that the number of the ways to first extract an element from the first set and then an element from the second set $= nm$ -- direct product of sets $X$ and $Y$.
${n+m}choose{2}$$2!= frac{(n+m)!}{2! (n+m-2)!} = frac{(n+m)(n+m-1)}{2}$ --the number of non-ordered pairs in set $Xcup Y$, where $|Xcup Y| = n + m$.
What the difference between these two formulas? I cannot see... Why $frac{(n+m)(n+m-1)}{2} neq nm$?
combinatorics
asked Dec 29 '18 at 0:02
Just do itJust do it
19618
$endgroup$
Product rule: if $X,Y$ -- finite sets and $|X|=n$, $|Y|= m$ we can say, that the number of the ways to first extract an element from the first set and then an element from the second set $= nm$ -- direct product of sets $X$ and $Y$.
${n+m}choose{2}$$2!= frac{(n+m)!}{2! (n+m-2)!} = frac{(n+m)(n+m-1)}{2}$ --the number of non-ordered pairs in set $Xcup Y$, where $|Xcup Y| = n + m$.
What the difference between these two formulas? I cannot see... Why $frac{(n+m)(n+m-1)}{2} neq nm$?
combinatorics
combinatorics
asked Dec 29 '18 at 0:02
Just do itJust do it
19618
asked Dec 29 '18 at 0:02
Just do itJust do it
19618
asked Dec 29 '18 at 0:02
Just do itJust do it
19618
asked Dec 29 '18 at 0:02
Just do itJust do it
19618
asked Dec 29 '18 at 0:02
Just do itJust do it
19618
19618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.
In the second scenario, there are three cases:
- Two elements from $X$
- Two elements from $Y$
- One element from $X$, one element from $Y$
This is why there are more pairs in the second scenario than the first.
answered Dec 29 '18 at 0:05
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
$begingroup$
Ou, oh, ye! Thx :D
$endgroup$
– Just do it
Dec 29 '18 at 0:06
1
$begingroup$
It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
$endgroup$
– Henry
Dec 29 '18 at 0:19
$begingroup$
@Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:38
$begingroup$
@Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:46
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%2f3055411%2fsimple-combinatorics-product-rule-and-number-of-ordered-pairs%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
$begingroup$
In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.
In the second scenario, there are three cases:
- Two elements from $X$
- Two elements from $Y$
- One element from $X$, one element from $Y$
This is why there are more pairs in the second scenario than the first.
answered Dec 29 '18 at 0:05
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
$begingroup$
Ou, oh, ye! Thx :D
$endgroup$
– Just do it
Dec 29 '18 at 0:06
1
$begingroup$
It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
$endgroup$
– Henry
Dec 29 '18 at 0:19
$begingroup$
@Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:38
$begingroup$
@Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:46
add a comment |
$begingroup$
In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.
In the second scenario, there are three cases:
- Two elements from $X$
- Two elements from $Y$
- One element from $X$, one element from $Y$
This is why there are more pairs in the second scenario than the first.
answered Dec 29 '18 at 0:05
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
$begingroup$
Ou, oh, ye! Thx :D
$endgroup$
– Just do it
Dec 29 '18 at 0:06
1
$begingroup$
It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
$endgroup$
– Henry
Dec 29 '18 at 0:19
$begingroup$
@Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:38
$begingroup$
@Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:46
add a comment |
$begingroup$
In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.
In the second scenario, there are three cases:
- Two elements from $X$
- Two elements from $Y$
- One element from $X$, one element from $Y$
This is why there are more pairs in the second scenario than the first.
answered Dec 29 '18 at 0:05
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
In the first scenario, there is one case: One element from $X$, one element from $Y$. That's the only type of pair you're going to get.
In the second scenario, there are three cases:
- Two elements from $X$
- Two elements from $Y$
- One element from $X$, one element from $Y$
This is why there are more pairs in the second scenario than the first.
answered Dec 29 '18 at 0:05
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 0:05
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 0:05
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 0:05
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
$begingroup$
Ou, oh, ye! Thx :D
$endgroup$
– Just do it
Dec 29 '18 at 0:06
1
$begingroup$
It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
$endgroup$
– Henry
Dec 29 '18 at 0:19
$begingroup$
@Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:38
$begingroup$
@Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:46
add a comment |
$begingroup$
Ou, oh, ye! Thx :D
$endgroup$
– Just do it
Dec 29 '18 at 0:06
1
$begingroup$
It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
$endgroup$
– Henry
Dec 29 '18 at 0:19
$begingroup$
@Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:38
$begingroup$
@Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:46
$begingroup$
Ou, oh, ye! Thx :D
$endgroup$
– Just do it
Dec 29 '18 at 0:06
$begingroup$
Ou, oh, ye! Thx :D
$endgroup$
– Just do it
Dec 29 '18 at 0:06
1
1
$begingroup$
It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
$endgroup$
– Henry
Dec 29 '18 at 0:19
$begingroup$
It might be worth saying that $dfrac{n(n-1)}{2} +dfrac{m(m-1)}{2} +nm = dfrac{(n+m)(n+m-1)}{2}$
$endgroup$
– Henry
Dec 29 '18 at 0:19
$begingroup$
@Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:38
$begingroup$
@Henry nice, $=$ ${n}choose{2}$ $+$ ${m}choose{2}$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:38
$begingroup$
@Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:46
$begingroup$
@Henry also clear, that if we use not direct product, but free product $X oplus X$, if $x_i in X$ we will be have two words in $Xoplus X$ $(x_i, x_i) neq (x_i, x_i)$ that equal in $X^2$ and in this formula we will be write ${n}choose{2}$$2!$ $+$ ${m}choose{2}$$2!$ $+$ $nm$
$endgroup$
– Just do it
Dec 29 '18 at 0:46
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%2f3055411%2fsimple-combinatorics-product-rule-and-number-of-ordered-pairs%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
