Inviting 5 friends out of 11 to dinner, with restriction
I'm revising for finals and I have come across this following question:
a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.
b) Repeat a) but 2 of the friends are married and will not attend separately.
For a) I got ${11 choose 5} = frac{11!}{(11-5)! cdot 5!} = 462$.
I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 choose 5} = frac{9!}{(9-5)! cdot 5!} = 126$, but I'm pretty sure this is wrong,
Could someone explain b) to me?
Thanks in advance!
probability combinatorics
edited Jun 14 '13 at 19:29
vadim123
75.7k897189
asked Apr 22 '12 at 14:54
XabiXabi
1713715
add a comment |
I'm revising for finals and I have come across this following question:
a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.
b) Repeat a) but 2 of the friends are married and will not attend separately.
For a) I got ${11 choose 5} = frac{11!}{(11-5)! cdot 5!} = 462$.
I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 choose 5} = frac{9!}{(9-5)! cdot 5!} = 126$, but I'm pretty sure this is wrong,
Could someone explain b) to me?
Thanks in advance!
probability combinatorics
edited Jun 14 '13 at 19:29
vadim123
75.7k897189
asked Apr 22 '12 at 14:54
XabiXabi
1713715
add a comment |
I'm revising for finals and I have come across this following question:
a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.
b) Repeat a) but 2 of the friends are married and will not attend separately.
For a) I got ${11 choose 5} = frac{11!}{(11-5)! cdot 5!} = 462$.
I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 choose 5} = frac{9!}{(9-5)! cdot 5!} = 126$, but I'm pretty sure this is wrong,
Could someone explain b) to me?
Thanks in advance!
probability combinatorics
edited Jun 14 '13 at 19:29
vadim123
75.7k897189
asked Apr 22 '12 at 14:54
XabiXabi
1713715
I'm revising for finals and I have come across this following question:
a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.
b) Repeat a) but 2 of the friends are married and will not attend separately.
For a) I got ${11 choose 5} = frac{11!}{(11-5)! cdot 5!} = 462$.
I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 choose 5} = frac{9!}{(9-5)! cdot 5!} = 126$, but I'm pretty sure this is wrong,
Could someone explain b) to me?
Thanks in advance!
probability combinatorics
probability combinatorics
edited Jun 14 '13 at 19:29
vadim123
75.7k897189
asked Apr 22 '12 at 14:54
XabiXabi
1713715
edited Jun 14 '13 at 19:29
vadim123
75.7k897189
asked Apr 22 '12 at 14:54
XabiXabi
1713715
edited Jun 14 '13 at 19:29
vadim123
75.7k897189
edited Jun 14 '13 at 19:29
vadim123
75.7k897189
edited Jun 14 '13 at 19:29
vadim123
75.7k897189
75.7k897189
asked Apr 22 '12 at 14:54
XabiXabi
1713715
asked Apr 22 '12 at 14:54
XabiXabi
1713715
asked Apr 22 '12 at 14:54
XabiXabi
1713715
1713715
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMMTMM
9,12032848
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
TomarinatorTomarinator
1,32111023
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%2f135322%2finviting-5-friends-out-of-11-to-dinner-with-restriction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMMTMM
9,12032848
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMMTMM
9,12032848
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMMTMM
9,12032848
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMMTMM
9,12032848
answered Apr 22 '12 at 15:02
TMMTMM
9,12032848
answered Apr 22 '12 at 15:02
TMMTMM
9,12032848
answered Apr 22 '12 at 15:02
TMMTMM
9,12032848
9,12032848
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
TomarinatorTomarinator
1,32111023
add a comment |
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
TomarinatorTomarinator
1,32111023
add a comment |
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
TomarinatorTomarinator
1,32111023
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
TomarinatorTomarinator
1,32111023
answered Apr 22 '12 at 15:06
TomarinatorTomarinator
1,32111023
answered Apr 22 '12 at 15:06
TomarinatorTomarinator
1,32111023
answered Apr 22 '12 at 15:06
TomarinatorTomarinator
1,32111023
1,32111023
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%2f135322%2finviting-5-friends-out-of-11-to-dinner-with-restriction%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
