How many ways are there to answer 10 questions with given conditions?Combination - How many different waysIn how many ways one can answer a 20-question exam with two-choice questions?How many ways are there to place the ball in N different bags under given condition?In how many ways can the students answer a 10-question true false examination?How many T/F answer keys are possible?Multiple choice exam where no three consecutive answers are the same (1)In how many ways can a candidate choose questions?How many candidates are appearing for exam?Combinatorics, how many ways can you select these essay questions?Why my process is wrong:-How many ways are there to choose $5$ questions from three sets of $4$, with at least one from each set?
Can you reject a postdoc offer after the PI has paid a large sum for flights/accommodation for your visit?
US to Europe trip with Canada layover- is 52 minutes enough?
Silly Sally's Movie
Deleting missing values from a dataset
Plywood subfloor won't screw down in a trailer home
Is having access to past exams cheating and, if yes, could it be proven just by a good grade?
How could a female member of a species produce eggs unto death?
Why doesn't the EU now just force the UK to choose between referendum and no-deal?
Best mythical creature to use as livestock?
How do anti-virus programs start at Windows boot?
Is it ok to include an epilogue dedicated to colleagues who passed away in the end of the manuscript?
What is the dot in “1.2.4."
It's a yearly task, alright
Latest web browser compatible with Windows 98
Running a subshell from the middle of the current command
How does Dispel Magic work against Stoneskin?
Playing ONE triplet (not three)
Humans have energy, but not water. What happens?
Does splitting a potentially monolithic application into several smaller ones help prevent bugs?
Does the Bracer of Flying Daggers benefit from the Dueling Fighting style?
Co-worker team leader wants to inject the crap software product of his friends into our development. What should I say to our common boss?
Why must traveling waves have the same amplitude to form a standing wave?
When is a batch class instantiated when you schedule it?
Do f-stop and exposure time perfectly cancel?
How many ways are there to answer 10 questions with given conditions?
Combination - How many different waysIn how many ways one can answer a 20-question exam with two-choice questions?How many ways are there to place the ball in N different bags under given condition?In how many ways can the students answer a 10-question true false examination?How many T/F answer keys are possible?Multiple choice exam where no three consecutive answers are the same (1)In how many ways can a candidate choose questions?How many candidates are appearing for exam?Combinatorics, how many ways can you select these essay questions?Why my process is wrong:-How many ways are there to choose $5$ questions from three sets of $4$, with at least one from each set?
$begingroup$
My question is this: there is an exam with 10 questions, they are all true or false questions. With the given two conditions how many ways are there to answer: There are no unanswered questions and no two consecutive problems are answered correctly.
combinatorics
asked Mar 10 at 19:29
bensubensu
52
$endgroup$
add a comment |
$begingroup$
My question is this: there is an exam with 10 questions, they are all true or false questions. With the given two conditions how many ways are there to answer: There are no unanswered questions and no two consecutive problems are answered correctly.
combinatorics
asked Mar 10 at 19:29
bensubensu
52
$endgroup$
add a comment |
$begingroup$
My question is this: there is an exam with 10 questions, they are all true or false questions. With the given two conditions how many ways are there to answer: There are no unanswered questions and no two consecutive problems are answered correctly.
combinatorics
asked Mar 10 at 19:29
bensubensu
52
$endgroup$
My question is this: there is an exam with 10 questions, they are all true or false questions. With the given two conditions how many ways are there to answer: There are no unanswered questions and no two consecutive problems are answered correctly.
combinatorics
combinatorics
asked Mar 10 at 19:29
bensubensu
52
asked Mar 10 at 19:29
bensubensu
52
asked Mar 10 at 19:29
bensubensu
52
asked Mar 10 at 19:29
bensubensu
52
asked Mar 10 at 19:29
bensubensu
52
52
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Presumably, the exam has already been written and the answer key is already set in stone. Rather than thinking of sequences of True / False, we should instead consider sequences of Correct ($colorgreencheckmark$) / Incorrect ($colorredtimes$) and count how many sequences of length ten there are with no two (or more) corrects in a row.
Consider how many ways there are to have $c$ questions correct and $10-c$ questions incorrect. To see this, consider placing the $10-c$ incorrects in a line and make a little space between and to the side of each in which we have the option of placing at most one checkmark between, noting that the order of doing so is irrelevant:
$$underbraceunderline~~~overbracecolorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes^10-c~~textspaces for colorredtimes's~underline~~_10-c+1~~textspaces available for ~colorgreencheckmark's$$
Simultaneously pick which spaces happen to be used by our $c$ checkmarks in $binom10-c+1c$ and then remove all unused empty spaces, leaving us with a sequence of ten checks or x's where no two checks are next to one another.
Range over all possible values of $c$ to get a final count (noting that $c$ can range from $0$ to $5$ since if there are $6$ or more correct answers there would necessarily be at least two adjacent):
$$sumlimits_c=0^5 binom10-c+1c = 144$$
I encourage you to try thinking about this problem in other ways as well.
Let $f(n)$ count the number of ways you could have such a sequence of corrects/incorrects of length $n$ with no two corrects in a row. By noting that such a sequence either begins with a correct or doesn't, and if it does begin with a correct it must be followed by an incorrect and then in either case what follows would need to be another valid sequence but shorter, you arrive at the following recurrence relation:
$$f(0)=1$$
$$f(1)=2$$
$$f(n)=f(n-1)+f(n-2)$$
which you should recognize as being the fibonacci sequence with an index shift. We find that $f(n)=F_n+2$ where $F_n$ is the fibonacci sequence.
Indeed, this coincides with what we found earlier, $f(10)=144 = F_12$ is the twelvth fibonacci number.
answered Mar 10 at 19:44
JMoravitzJMoravitz
48.5k33988
$endgroup$
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%2f3142770%2fhow-many-ways-are-there-to-answer-10-questions-with-given-conditions%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
$begingroup$
Presumably, the exam has already been written and the answer key is already set in stone. Rather than thinking of sequences of True / False, we should instead consider sequences of Correct ($colorgreencheckmark$) / Incorrect ($colorredtimes$) and count how many sequences of length ten there are with no two (or more) corrects in a row.
Consider how many ways there are to have $c$ questions correct and $10-c$ questions incorrect. To see this, consider placing the $10-c$ incorrects in a line and make a little space between and to the side of each in which we have the option of placing at most one checkmark between, noting that the order of doing so is irrelevant:
$$underbraceunderline~~~overbracecolorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes^10-c~~textspaces for colorredtimes's~underline~~_10-c+1~~textspaces available for ~colorgreencheckmark's$$
Simultaneously pick which spaces happen to be used by our $c$ checkmarks in $binom10-c+1c$ and then remove all unused empty spaces, leaving us with a sequence of ten checks or x's where no two checks are next to one another.
Range over all possible values of $c$ to get a final count (noting that $c$ can range from $0$ to $5$ since if there are $6$ or more correct answers there would necessarily be at least two adjacent):
$$sumlimits_c=0^5 binom10-c+1c = 144$$
I encourage you to try thinking about this problem in other ways as well.
Let $f(n)$ count the number of ways you could have such a sequence of corrects/incorrects of length $n$ with no two corrects in a row. By noting that such a sequence either begins with a correct or doesn't, and if it does begin with a correct it must be followed by an incorrect and then in either case what follows would need to be another valid sequence but shorter, you arrive at the following recurrence relation:
$$f(0)=1$$
$$f(1)=2$$
$$f(n)=f(n-1)+f(n-2)$$
which you should recognize as being the fibonacci sequence with an index shift. We find that $f(n)=F_n+2$ where $F_n$ is the fibonacci sequence.
Indeed, this coincides with what we found earlier, $f(10)=144 = F_12$ is the twelvth fibonacci number.
answered Mar 10 at 19:44
JMoravitzJMoravitz
48.5k33988
$endgroup$
add a comment |
$begingroup$
Presumably, the exam has already been written and the answer key is already set in stone. Rather than thinking of sequences of True / False, we should instead consider sequences of Correct ($colorgreencheckmark$) / Incorrect ($colorredtimes$) and count how many sequences of length ten there are with no two (or more) corrects in a row.
Consider how many ways there are to have $c$ questions correct and $10-c$ questions incorrect. To see this, consider placing the $10-c$ incorrects in a line and make a little space between and to the side of each in which we have the option of placing at most one checkmark between, noting that the order of doing so is irrelevant:
$$underbraceunderline~~~overbracecolorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes^10-c~~textspaces for colorredtimes's~underline~~_10-c+1~~textspaces available for ~colorgreencheckmark's$$
Simultaneously pick which spaces happen to be used by our $c$ checkmarks in $binom10-c+1c$ and then remove all unused empty spaces, leaving us with a sequence of ten checks or x's where no two checks are next to one another.
Range over all possible values of $c$ to get a final count (noting that $c$ can range from $0$ to $5$ since if there are $6$ or more correct answers there would necessarily be at least two adjacent):
$$sumlimits_c=0^5 binom10-c+1c = 144$$
I encourage you to try thinking about this problem in other ways as well.
Let $f(n)$ count the number of ways you could have such a sequence of corrects/incorrects of length $n$ with no two corrects in a row. By noting that such a sequence either begins with a correct or doesn't, and if it does begin with a correct it must be followed by an incorrect and then in either case what follows would need to be another valid sequence but shorter, you arrive at the following recurrence relation:
$$f(0)=1$$
$$f(1)=2$$
$$f(n)=f(n-1)+f(n-2)$$
which you should recognize as being the fibonacci sequence with an index shift. We find that $f(n)=F_n+2$ where $F_n$ is the fibonacci sequence.
Indeed, this coincides with what we found earlier, $f(10)=144 = F_12$ is the twelvth fibonacci number.
answered Mar 10 at 19:44
JMoravitzJMoravitz
48.5k33988
$endgroup$
add a comment |
$begingroup$
Presumably, the exam has already been written and the answer key is already set in stone. Rather than thinking of sequences of True / False, we should instead consider sequences of Correct ($colorgreencheckmark$) / Incorrect ($colorredtimes$) and count how many sequences of length ten there are with no two (or more) corrects in a row.
Consider how many ways there are to have $c$ questions correct and $10-c$ questions incorrect. To see this, consider placing the $10-c$ incorrects in a line and make a little space between and to the side of each in which we have the option of placing at most one checkmark between, noting that the order of doing so is irrelevant:
$$underbraceunderline~~~overbracecolorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes^10-c~~textspaces for colorredtimes's~underline~~_10-c+1~~textspaces available for ~colorgreencheckmark's$$
Simultaneously pick which spaces happen to be used by our $c$ checkmarks in $binom10-c+1c$ and then remove all unused empty spaces, leaving us with a sequence of ten checks or x's where no two checks are next to one another.
Range over all possible values of $c$ to get a final count (noting that $c$ can range from $0$ to $5$ since if there are $6$ or more correct answers there would necessarily be at least two adjacent):
$$sumlimits_c=0^5 binom10-c+1c = 144$$
I encourage you to try thinking about this problem in other ways as well.
Let $f(n)$ count the number of ways you could have such a sequence of corrects/incorrects of length $n$ with no two corrects in a row. By noting that such a sequence either begins with a correct or doesn't, and if it does begin with a correct it must be followed by an incorrect and then in either case what follows would need to be another valid sequence but shorter, you arrive at the following recurrence relation:
$$f(0)=1$$
$$f(1)=2$$
$$f(n)=f(n-1)+f(n-2)$$
which you should recognize as being the fibonacci sequence with an index shift. We find that $f(n)=F_n+2$ where $F_n$ is the fibonacci sequence.
Indeed, this coincides with what we found earlier, $f(10)=144 = F_12$ is the twelvth fibonacci number.
answered Mar 10 at 19:44
JMoravitzJMoravitz
48.5k33988
$endgroup$
Presumably, the exam has already been written and the answer key is already set in stone. Rather than thinking of sequences of True / False, we should instead consider sequences of Correct ($colorgreencheckmark$) / Incorrect ($colorredtimes$) and count how many sequences of length ten there are with no two (or more) corrects in a row.
Consider how many ways there are to have $c$ questions correct and $10-c$ questions incorrect. To see this, consider placing the $10-c$ incorrects in a line and make a little space between and to the side of each in which we have the option of placing at most one checkmark between, noting that the order of doing so is irrelevant:
$$underbraceunderline~~~overbracecolorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes~underline~~~colorredtimes^10-c~~textspaces for colorredtimes's~underline~~_10-c+1~~textspaces available for ~colorgreencheckmark's$$
Simultaneously pick which spaces happen to be used by our $c$ checkmarks in $binom10-c+1c$ and then remove all unused empty spaces, leaving us with a sequence of ten checks or x's where no two checks are next to one another.
Range over all possible values of $c$ to get a final count (noting that $c$ can range from $0$ to $5$ since if there are $6$ or more correct answers there would necessarily be at least two adjacent):
$$sumlimits_c=0^5 binom10-c+1c = 144$$
I encourage you to try thinking about this problem in other ways as well.
Let $f(n)$ count the number of ways you could have such a sequence of corrects/incorrects of length $n$ with no two corrects in a row. By noting that such a sequence either begins with a correct or doesn't, and if it does begin with a correct it must be followed by an incorrect and then in either case what follows would need to be another valid sequence but shorter, you arrive at the following recurrence relation:
$$f(0)=1$$
$$f(1)=2$$
$$f(n)=f(n-1)+f(n-2)$$
which you should recognize as being the fibonacci sequence with an index shift. We find that $f(n)=F_n+2$ where $F_n$ is the fibonacci sequence.
Indeed, this coincides with what we found earlier, $f(10)=144 = F_12$ is the twelvth fibonacci number.
answered Mar 10 at 19:44
JMoravitzJMoravitz
48.5k33988
answered Mar 10 at 19:44
JMoravitzJMoravitz
48.5k33988
answered Mar 10 at 19:44
JMoravitzJMoravitz
48.5k33988
answered Mar 10 at 19:44
JMoravitzJMoravitz
48.5k33988
48.5k33988
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%2f3142770%2fhow-many-ways-are-there-to-answer-10-questions-with-given-conditions%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
