Mixing
Picking exactly $3$ different suits in a $4$ card hand
$begingroup$
Given a standard $52$ card deck, I want to know how many different ways it is possible to pick four cards so that you have exactly $3$ suits in your hand (i.e. there is exactly one suit pair). Additionally, order does matter, so for example: ace of hearts, two of hearts, ace of clubs, ace of spades is distinct from two of hearts, ace of hearts, ace of clubs, ace of spades.
I have two approaches to this problem, but I can't figure out which (or either) is right.
1) $52$ cards to start with. Then $13$ possible cards of one suit, $13$ possible cards of another suit, $12$ possible cards of whichever suit already have. $52 cdot 13 cdot 13 cdot 12 = 105,456$
2) $4!$ combinations. $13$ choose $2$ of one suit, $13$ choose $1$ of another, $13$ choose $1$ of another. $4! cdot binom{13}{2}binom{13}{1}binom{13}{1} = 316,368$
Thanks!
combinatorics
edited Jan 14 at 11:09
N. F. Taussig
44.2k93356
asked Jan 14 at 2:25
appleapple
445
$endgroup$
add a comment |
$begingroup$
Given a standard $52$ card deck, I want to know how many different ways it is possible to pick four cards so that you have exactly $3$ suits in your hand (i.e. there is exactly one suit pair). Additionally, order does matter, so for example: ace of hearts, two of hearts, ace of clubs, ace of spades is distinct from two of hearts, ace of hearts, ace of clubs, ace of spades.
I have two approaches to this problem, but I can't figure out which (or either) is right.
1) $52$ cards to start with. Then $13$ possible cards of one suit, $13$ possible cards of another suit, $12$ possible cards of whichever suit already have. $52 cdot 13 cdot 13 cdot 12 = 105,456$
2) $4!$ combinations. $13$ choose $2$ of one suit, $13$ choose $1$ of another, $13$ choose $1$ of another. $4! cdot binom{13}{2}binom{13}{1}binom{13}{1} = 316,368$
Thanks!
combinatorics
edited Jan 14 at 11:09
N. F. Taussig
44.2k93356
asked Jan 14 at 2:25
appleapple
445
$endgroup$
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 14 at 11:15
add a comment |
$begingroup$
Given a standard $52$ card deck, I want to know how many different ways it is possible to pick four cards so that you have exactly $3$ suits in your hand (i.e. there is exactly one suit pair). Additionally, order does matter, so for example: ace of hearts, two of hearts, ace of clubs, ace of spades is distinct from two of hearts, ace of hearts, ace of clubs, ace of spades.
I have two approaches to this problem, but I can't figure out which (or either) is right.
1) $52$ cards to start with. Then $13$ possible cards of one suit, $13$ possible cards of another suit, $12$ possible cards of whichever suit already have. $52 cdot 13 cdot 13 cdot 12 = 105,456$
2) $4!$ combinations. $13$ choose $2$ of one suit, $13$ choose $1$ of another, $13$ choose $1$ of another. $4! cdot binom{13}{2}binom{13}{1}binom{13}{1} = 316,368$
Thanks!
combinatorics
edited Jan 14 at 11:09
N. F. Taussig
44.2k93356
asked Jan 14 at 2:25
appleapple
445
$endgroup$
Given a standard $52$ card deck, I want to know how many different ways it is possible to pick four cards so that you have exactly $3$ suits in your hand (i.e. there is exactly one suit pair). Additionally, order does matter, so for example: ace of hearts, two of hearts, ace of clubs, ace of spades is distinct from two of hearts, ace of hearts, ace of clubs, ace of spades.
I have two approaches to this problem, but I can't figure out which (or either) is right.
1) $52$ cards to start with. Then $13$ possible cards of one suit, $13$ possible cards of another suit, $12$ possible cards of whichever suit already have. $52 cdot 13 cdot 13 cdot 12 = 105,456$
2) $4!$ combinations. $13$ choose $2$ of one suit, $13$ choose $1$ of another, $13$ choose $1$ of another. $4! cdot binom{13}{2}binom{13}{1}binom{13}{1} = 316,368$
Thanks!
combinatorics
combinatorics
edited Jan 14 at 11:09
N. F. Taussig
44.2k93356
asked Jan 14 at 2:25
appleapple
445
edited Jan 14 at 11:09
N. F. Taussig
44.2k93356
asked Jan 14 at 2:25
appleapple
445
edited Jan 14 at 11:09
N. F. Taussig
44.2k93356
edited Jan 14 at 11:09
N. F. Taussig
44.2k93356
edited Jan 14 at 11:09
N. F. Taussig
44.2k93356
44.2k93356
asked Jan 14 at 2:25
appleapple
445
asked Jan 14 at 2:25
appleapple
445
asked Jan 14 at 2:25
appleapple
445
445
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 14 at 11:15
add a comment |
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 14 at 11:15
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 14 at 11:15
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 14 at 11:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are four ways to select the suit from which two cards are drawn, $binom{13}{2}$ ways to select two of the thirteen cards of that suit, $binom{3}{2}$ ways to select two of the remaining three suits, thirteen ways to select a card from each of those suits, and $4!$ ways to arrange the selected cards.
$$binom{4}{1}binom{13}{2}binom{3}{2}binom{13}{1}^24!$$
Your first approach is wrong since the second card of the same suit may be picked with the second, third, or fourth card. Also, you have not selected from which suits the cards will be drawn.
Your second approach is wrong since you have not selected the suits from which the cards will be drawn.
answered Jan 14 at 2:33
N. F. TaussigN. F. Taussig
44.2k93356
$endgroup$
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%2f3072782%2fpicking-exactly-3-different-suits-in-a-4-card-hand%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
$begingroup$
There are four ways to select the suit from which two cards are drawn, $binom{13}{2}$ ways to select two of the thirteen cards of that suit, $binom{3}{2}$ ways to select two of the remaining three suits, thirteen ways to select a card from each of those suits, and $4!$ ways to arrange the selected cards.
$$binom{4}{1}binom{13}{2}binom{3}{2}binom{13}{1}^24!$$
Your first approach is wrong since the second card of the same suit may be picked with the second, third, or fourth card. Also, you have not selected from which suits the cards will be drawn.
Your second approach is wrong since you have not selected the suits from which the cards will be drawn.
answered Jan 14 at 2:33
N. F. TaussigN. F. Taussig
44.2k93356
$endgroup$
add a comment |
$begingroup$
There are four ways to select the suit from which two cards are drawn, $binom{13}{2}$ ways to select two of the thirteen cards of that suit, $binom{3}{2}$ ways to select two of the remaining three suits, thirteen ways to select a card from each of those suits, and $4!$ ways to arrange the selected cards.
$$binom{4}{1}binom{13}{2}binom{3}{2}binom{13}{1}^24!$$
Your first approach is wrong since the second card of the same suit may be picked with the second, third, or fourth card. Also, you have not selected from which suits the cards will be drawn.
Your second approach is wrong since you have not selected the suits from which the cards will be drawn.
answered Jan 14 at 2:33
N. F. TaussigN. F. Taussig
44.2k93356
$endgroup$
add a comment |
$begingroup$
There are four ways to select the suit from which two cards are drawn, $binom{13}{2}$ ways to select two of the thirteen cards of that suit, $binom{3}{2}$ ways to select two of the remaining three suits, thirteen ways to select a card from each of those suits, and $4!$ ways to arrange the selected cards.
$$binom{4}{1}binom{13}{2}binom{3}{2}binom{13}{1}^24!$$
Your first approach is wrong since the second card of the same suit may be picked with the second, third, or fourth card. Also, you have not selected from which suits the cards will be drawn.
Your second approach is wrong since you have not selected the suits from which the cards will be drawn.
answered Jan 14 at 2:33
N. F. TaussigN. F. Taussig
44.2k93356
$endgroup$
There are four ways to select the suit from which two cards are drawn, $binom{13}{2}$ ways to select two of the thirteen cards of that suit, $binom{3}{2}$ ways to select two of the remaining three suits, thirteen ways to select a card from each of those suits, and $4!$ ways to arrange the selected cards.
$$binom{4}{1}binom{13}{2}binom{3}{2}binom{13}{1}^24!$$
Your first approach is wrong since the second card of the same suit may be picked with the second, third, or fourth card. Also, you have not selected from which suits the cards will be drawn.
Your second approach is wrong since you have not selected the suits from which the cards will be drawn.
answered Jan 14 at 2:33
N. F. TaussigN. F. Taussig
44.2k93356
answered Jan 14 at 2:33
N. F. TaussigN. F. Taussig
44.2k93356
answered Jan 14 at 2:33
N. F. TaussigN. F. Taussig
44.2k93356
answered Jan 14 at 2:33
N. F. TaussigN. F. Taussig
44.2k93356
44.2k93356
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%2f3072782%2fpicking-exactly-3-different-suits-in-a-4-card-hand%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
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 14 at 11:15