Coins to be put in empty purses
$begingroup$
There are six empty purses on the table. What is the number of ways to put 12 identical 1 rupee coin in purse such that at most one purse remains empty.
The way I did is
Case 1: No purse remains empty.
In that case the answer should be $binom{11}{5}$
Case 2: One purse remains empty.
In this case answer should be
$6cdot binom{11}{4}$
So the answer should be
$binom{11}{5} + 6cdot binom{11}{4}$
But the answer given is
$binom{12}{5} + 6cdot binom{11}{4}$
I believe the answer is wrong.
Kindly verify.
combinatorics binomial-coefficients
edited Dec 22 '18 at 12:24
Shaun
9,366113684
asked Mar 21 '18 at 12:11
Akash GautamaAkash Gautama
885
$endgroup$
add a comment |
$begingroup$
There are six empty purses on the table. What is the number of ways to put 12 identical 1 rupee coin in purse such that at most one purse remains empty.
The way I did is
Case 1: No purse remains empty.
In that case the answer should be $binom{11}{5}$
Case 2: One purse remains empty.
In this case answer should be
$6cdot binom{11}{4}$
So the answer should be
$binom{11}{5} + 6cdot binom{11}{4}$
But the answer given is
$binom{12}{5} + 6cdot binom{11}{4}$
I believe the answer is wrong.
Kindly verify.
combinatorics binomial-coefficients
edited Dec 22 '18 at 12:24
Shaun
9,366113684
asked Mar 21 '18 at 12:11
Akash GautamaAkash Gautama
885
$endgroup$
$begingroup$
I understood the concept
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:41
$begingroup$
$binom{11}{5} + 6cdot binom{11}{4}$ = $binom{11}{5} + binom{11}{4}$ +5cdot binom{11}{4}$
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:42
$begingroup$
Right, it should $(6)$ times, not $(5)$ times (i.e., choose the empty purse). With that edit, your answer is correct.
$endgroup$
– quasi
Mar 21 '18 at 12:49
add a comment |
$begingroup$
There are six empty purses on the table. What is the number of ways to put 12 identical 1 rupee coin in purse such that at most one purse remains empty.
The way I did is
Case 1: No purse remains empty.
In that case the answer should be $binom{11}{5}$
Case 2: One purse remains empty.
In this case answer should be
$6cdot binom{11}{4}$
So the answer should be
$binom{11}{5} + 6cdot binom{11}{4}$
But the answer given is
$binom{12}{5} + 6cdot binom{11}{4}$
I believe the answer is wrong.
Kindly verify.
combinatorics binomial-coefficients
edited Dec 22 '18 at 12:24
Shaun
9,366113684
asked Mar 21 '18 at 12:11
Akash GautamaAkash Gautama
885
$endgroup$
There are six empty purses on the table. What is the number of ways to put 12 identical 1 rupee coin in purse such that at most one purse remains empty.
The way I did is
Case 1: No purse remains empty.
In that case the answer should be $binom{11}{5}$
Case 2: One purse remains empty.
In this case answer should be
$6cdot binom{11}{4}$
So the answer should be
$binom{11}{5} + 6cdot binom{11}{4}$
But the answer given is
$binom{12}{5} + 6cdot binom{11}{4}$
I believe the answer is wrong.
Kindly verify.
combinatorics binomial-coefficients
combinatorics binomial-coefficients
edited Dec 22 '18 at 12:24
Shaun
9,366113684
asked Mar 21 '18 at 12:11
Akash GautamaAkash Gautama
885
edited Dec 22 '18 at 12:24
Shaun
9,366113684
asked Mar 21 '18 at 12:11
Akash GautamaAkash Gautama
885
edited Dec 22 '18 at 12:24
Shaun
9,366113684
edited Dec 22 '18 at 12:24
Shaun
9,366113684
edited Dec 22 '18 at 12:24
Shaun
9,366113684
9,366113684
asked Mar 21 '18 at 12:11
Akash GautamaAkash Gautama
885
asked Mar 21 '18 at 12:11
Akash GautamaAkash Gautama
885
asked Mar 21 '18 at 12:11
Akash GautamaAkash Gautama
885
885
$begingroup$
I understood the concept
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:41
$begingroup$
$binom{11}{5} + 6cdot binom{11}{4}$ = $binom{11}{5} + binom{11}{4}$ +5cdot binom{11}{4}$
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:42
$begingroup$
Right, it should $(6)$ times, not $(5)$ times (i.e., choose the empty purse). With that edit, your answer is correct.
$endgroup$
– quasi
Mar 21 '18 at 12:49
add a comment |
$begingroup$
I understood the concept
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:41
$begingroup$
$binom{11}{5} + 6cdot binom{11}{4}$ = $binom{11}{5} + binom{11}{4}$ +5cdot binom{11}{4}$
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:42
$begingroup$
Right, it should $(6)$ times, not $(5)$ times (i.e., choose the empty purse). With that edit, your answer is correct.
$endgroup$
– quasi
Mar 21 '18 at 12:49
$begingroup$
I understood the concept
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:41
$begingroup$
I understood the concept
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:41
$begingroup$
$binom{11}{5} + 6cdot binom{11}{4}$ = $binom{11}{5} + binom{11}{4}$ +5cdot binom{11}{4}$
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:42
$begingroup$
$binom{11}{5} + 6cdot binom{11}{4}$ = $binom{11}{5} + binom{11}{4}$ +5cdot binom{11}{4}$
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:42
$begingroup$
Right, it should $(6)$ times, not $(5)$ times (i.e., choose the empty purse). With that edit, your answer is correct.
$endgroup$
– quasi
Mar 21 '18 at 12:49
$begingroup$
Right, it should $(6)$ times, not $(5)$ times (i.e., choose the empty purse). With that edit, your answer is correct.
$endgroup$
– quasi
Mar 21 '18 at 12:49
add a comment |
0
active
oldest
votes
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%2f2701671%2fcoins-to-be-put-in-empty-purses%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2701671%2fcoins-to-be-put-in-empty-purses%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
$begingroup$
I understood the concept
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:41
$begingroup$
$binom{11}{5} + 6cdot binom{11}{4}$ = $binom{11}{5} + binom{11}{4}$ +5cdot binom{11}{4}$
$endgroup$
– Akash Gautama
Mar 21 '18 at 12:42
$begingroup$
Right, it should $(6)$ times, not $(5)$ times (i.e., choose the empty purse). With that edit, your answer is correct.
$endgroup$
– quasi
Mar 21 '18 at 12:49