Permutation and combination Number of Bits Problem [closed]
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how many bit strings of length eight either start with a '1' bit or end with the two bits '00'?
I got the answer 128 by using the "exclusive or" approach i.e. for bit strings starting with '1': There is one way to choose the first bit, 2^5 ways to choose the succeeding 5 bits and 3 ways to choose the last 2 bits= 3*2^5
and for bits ending with '00': There is one way to choose the first and last two bits and 2^5 ways to choose the remaining bits= 2^5
Therefore answer will be= 3*2^5+2^5=128.
combinatorics discrete-mathematics logic permutations combinations
asked Dec 17 '18 at 13:33
Apoorv GargApoorv Garg
11
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closed as off-topic by amWhy, Saad, Lord Shark the Unknown, mrtaurho, Ben Dec 18 '18 at 7:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Saad, mrtaurho, Ben
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
how many bit strings of length eight either start with a '1' bit or end with the two bits '00'?
I got the answer 128 by using the "exclusive or" approach i.e. for bit strings starting with '1': There is one way to choose the first bit, 2^5 ways to choose the succeeding 5 bits and 3 ways to choose the last 2 bits= 3*2^5
and for bits ending with '00': There is one way to choose the first and last two bits and 2^5 ways to choose the remaining bits= 2^5
Therefore answer will be= 3*2^5+2^5=128.
combinatorics discrete-mathematics logic permutations combinations
asked Dec 17 '18 at 13:33
Apoorv GargApoorv Garg
11
$endgroup$
closed as off-topic by amWhy, Saad, Lord Shark the Unknown, mrtaurho, Ben Dec 18 '18 at 7:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Saad, mrtaurho, Ben
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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Assuming you meant the non-exclusive or: There are $2^7$ that begin with $1$. There are $2^6$ that end in $00$. There are $2^5$ that BOTH begin with $1$ and end in $00$. Thus the answer is $2^7+2^6-2^5=160$. Assuming you meant the exclusive or: we then have to subtract $2^5$ again, now getting your $128$. So...did you mean the exclusive or? That should be specified.
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– lulu
Dec 17 '18 at 13:45
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@lulu I noticed that you put you answerS as comments. Why? (just curious)
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– adhg
Dec 18 '18 at 1:11
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Can you add how you got your answer of $128$? We'll be able to help you better if we can see how you got your answer.
$endgroup$
– Carl Schildkraut
Dec 18 '18 at 4:26
1
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@adhg That was hardly an answer...I was just pointing out that the question was ambiguous and that (what I imagine to be) the standard reading would yield a different answer but that a (somewhat) non-standard reading would give the "right" answer. I was asking the OP for clarification.
$endgroup$
– lulu
Dec 18 '18 at 12:00
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I think "Either" in this question signify 'Exclusive or' itself and there is no need to specify it.
$endgroup$
– Apoorv Garg
Dec 19 '18 at 8:07
add a comment |
$begingroup$
how many bit strings of length eight either start with a '1' bit or end with the two bits '00'?
I got the answer 128 by using the "exclusive or" approach i.e. for bit strings starting with '1': There is one way to choose the first bit, 2^5 ways to choose the succeeding 5 bits and 3 ways to choose the last 2 bits= 3*2^5
and for bits ending with '00': There is one way to choose the first and last two bits and 2^5 ways to choose the remaining bits= 2^5
Therefore answer will be= 3*2^5+2^5=128.
combinatorics discrete-mathematics logic permutations combinations
asked Dec 17 '18 at 13:33
Apoorv GargApoorv Garg
11
$endgroup$
how many bit strings of length eight either start with a '1' bit or end with the two bits '00'?
I got the answer 128 by using the "exclusive or" approach i.e. for bit strings starting with '1': There is one way to choose the first bit, 2^5 ways to choose the succeeding 5 bits and 3 ways to choose the last 2 bits= 3*2^5
and for bits ending with '00': There is one way to choose the first and last two bits and 2^5 ways to choose the remaining bits= 2^5
Therefore answer will be= 3*2^5+2^5=128.
combinatorics discrete-mathematics logic permutations combinations
combinatorics discrete-mathematics logic permutations combinations
asked Dec 17 '18 at 13:33
Apoorv GargApoorv Garg
11
asked Dec 17 '18 at 13:33
Apoorv GargApoorv Garg
11
edited Dec 19 '18 at 9:55
Apoorv Garg
asked Dec 17 '18 at 13:33
Apoorv GargApoorv Garg
11
asked Dec 17 '18 at 13:33
Apoorv GargApoorv Garg
11
asked Dec 17 '18 at 13:33
Apoorv GargApoorv Garg
11
11
closed as off-topic by amWhy, Saad, Lord Shark the Unknown, mrtaurho, Ben Dec 18 '18 at 7:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Saad, mrtaurho, Ben
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Saad, Lord Shark the Unknown, mrtaurho, Ben Dec 18 '18 at 7:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Saad, mrtaurho, Ben
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Assuming you meant the non-exclusive or: There are $2^7$ that begin with $1$. There are $2^6$ that end in $00$. There are $2^5$ that BOTH begin with $1$ and end in $00$. Thus the answer is $2^7+2^6-2^5=160$. Assuming you meant the exclusive or: we then have to subtract $2^5$ again, now getting your $128$. So...did you mean the exclusive or? That should be specified.
$endgroup$
– lulu
Dec 17 '18 at 13:45
$begingroup$
@lulu I noticed that you put you answerS as comments. Why? (just curious)
$endgroup$
– adhg
Dec 18 '18 at 1:11
$begingroup$
Can you add how you got your answer of $128$? We'll be able to help you better if we can see how you got your answer.
$endgroup$
– Carl Schildkraut
Dec 18 '18 at 4:26
1
$begingroup$
@adhg That was hardly an answer...I was just pointing out that the question was ambiguous and that (what I imagine to be) the standard reading would yield a different answer but that a (somewhat) non-standard reading would give the "right" answer. I was asking the OP for clarification.
$endgroup$
– lulu
Dec 18 '18 at 12:00
$begingroup$
I think "Either" in this question signify 'Exclusive or' itself and there is no need to specify it.
$endgroup$
– Apoorv Garg
Dec 19 '18 at 8:07
add a comment |
2
$begingroup$
Assuming you meant the non-exclusive or: There are $2^7$ that begin with $1$. There are $2^6$ that end in $00$. There are $2^5$ that BOTH begin with $1$ and end in $00$. Thus the answer is $2^7+2^6-2^5=160$. Assuming you meant the exclusive or: we then have to subtract $2^5$ again, now getting your $128$. So...did you mean the exclusive or? That should be specified.
$endgroup$
– lulu
Dec 17 '18 at 13:45
$begingroup$
@lulu I noticed that you put you answerS as comments. Why? (just curious)
$endgroup$
– adhg
Dec 18 '18 at 1:11
$begingroup$
Can you add how you got your answer of $128$? We'll be able to help you better if we can see how you got your answer.
$endgroup$
– Carl Schildkraut
Dec 18 '18 at 4:26
1
$begingroup$
@adhg That was hardly an answer...I was just pointing out that the question was ambiguous and that (what I imagine to be) the standard reading would yield a different answer but that a (somewhat) non-standard reading would give the "right" answer. I was asking the OP for clarification.
$endgroup$
– lulu
Dec 18 '18 at 12:00
$begingroup$
I think "Either" in this question signify 'Exclusive or' itself and there is no need to specify it.
$endgroup$
– Apoorv Garg
Dec 19 '18 at 8:07
2
2
$begingroup$
Assuming you meant the non-exclusive or: There are $2^7$ that begin with $1$. There are $2^6$ that end in $00$. There are $2^5$ that BOTH begin with $1$ and end in $00$. Thus the answer is $2^7+2^6-2^5=160$. Assuming you meant the exclusive or: we then have to subtract $2^5$ again, now getting your $128$. So...did you mean the exclusive or? That should be specified.
$endgroup$
– lulu
Dec 17 '18 at 13:45
$begingroup$
Assuming you meant the non-exclusive or: There are $2^7$ that begin with $1$. There are $2^6$ that end in $00$. There are $2^5$ that BOTH begin with $1$ and end in $00$. Thus the answer is $2^7+2^6-2^5=160$. Assuming you meant the exclusive or: we then have to subtract $2^5$ again, now getting your $128$. So...did you mean the exclusive or? That should be specified.
$endgroup$
– lulu
Dec 17 '18 at 13:45
$begingroup$
@lulu I noticed that you put you answerS as comments. Why? (just curious)
$endgroup$
– adhg
Dec 18 '18 at 1:11
$begingroup$
@lulu I noticed that you put you answerS as comments. Why? (just curious)
$endgroup$
– adhg
Dec 18 '18 at 1:11
$begingroup$
Can you add how you got your answer of $128$? We'll be able to help you better if we can see how you got your answer.
$endgroup$
– Carl Schildkraut
Dec 18 '18 at 4:26
$begingroup$
Can you add how you got your answer of $128$? We'll be able to help you better if we can see how you got your answer.
$endgroup$
– Carl Schildkraut
Dec 18 '18 at 4:26
1
1
$begingroup$
@adhg That was hardly an answer...I was just pointing out that the question was ambiguous and that (what I imagine to be) the standard reading would yield a different answer but that a (somewhat) non-standard reading would give the "right" answer. I was asking the OP for clarification.
$endgroup$
– lulu
Dec 18 '18 at 12:00
$begingroup$
@adhg That was hardly an answer...I was just pointing out that the question was ambiguous and that (what I imagine to be) the standard reading would yield a different answer but that a (somewhat) non-standard reading would give the "right" answer. I was asking the OP for clarification.
$endgroup$
– lulu
Dec 18 '18 at 12:00
$begingroup$
I think "Either" in this question signify 'Exclusive or' itself and there is no need to specify it.
$endgroup$
– Apoorv Garg
Dec 19 '18 at 8:07
$begingroup$
I think "Either" in this question signify 'Exclusive or' itself and there is no need to specify it.
$endgroup$
– Apoorv Garg
Dec 19 '18 at 8:07
add a comment |
1 Answer
1
active
oldest
votes
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Recall that $$n(Acup B)=n(A)+n(B)-n(Acap B)$$
where $A$ is the set of $8$ length bit strings starting with $1,B$ is the set of $8$ length bit strings ending in $00$.
$n(A)=2^7$, because the remaining last $7$ bits can be $0,1$.
$n(B)=2^6$, because the remaining first $4$ bits can be $0,1$.
$n(Acap B)$ is the number of $8$ length bit strings which both start with $1$ and end in $00$. Thus, there are $5$ bits which we can vary, resulting in $n(Acap B)=2^5$.
The required answer is $2^7+2^6-2^5=128+64-32=160$.
answered Dec 17 '18 at 13:46
Shubham JohriShubham Johri
5,189718
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Recall that $$n(Acup B)=n(A)+n(B)-n(Acap B)$$
where $A$ is the set of $8$ length bit strings starting with $1,B$ is the set of $8$ length bit strings ending in $00$.
$n(A)=2^7$, because the remaining last $7$ bits can be $0,1$.
$n(B)=2^6$, because the remaining first $4$ bits can be $0,1$.
$n(Acap B)$ is the number of $8$ length bit strings which both start with $1$ and end in $00$. Thus, there are $5$ bits which we can vary, resulting in $n(Acap B)=2^5$.
The required answer is $2^7+2^6-2^5=128+64-32=160$.
answered Dec 17 '18 at 13:46
Shubham JohriShubham Johri
5,189718
$endgroup$
add a comment |
$begingroup$
Recall that $$n(Acup B)=n(A)+n(B)-n(Acap B)$$
where $A$ is the set of $8$ length bit strings starting with $1,B$ is the set of $8$ length bit strings ending in $00$.
$n(A)=2^7$, because the remaining last $7$ bits can be $0,1$.
$n(B)=2^6$, because the remaining first $4$ bits can be $0,1$.
$n(Acap B)$ is the number of $8$ length bit strings which both start with $1$ and end in $00$. Thus, there are $5$ bits which we can vary, resulting in $n(Acap B)=2^5$.
The required answer is $2^7+2^6-2^5=128+64-32=160$.
answered Dec 17 '18 at 13:46
Shubham JohriShubham Johri
5,189718
$endgroup$
add a comment |
$begingroup$
Recall that $$n(Acup B)=n(A)+n(B)-n(Acap B)$$
where $A$ is the set of $8$ length bit strings starting with $1,B$ is the set of $8$ length bit strings ending in $00$.
$n(A)=2^7$, because the remaining last $7$ bits can be $0,1$.
$n(B)=2^6$, because the remaining first $4$ bits can be $0,1$.
$n(Acap B)$ is the number of $8$ length bit strings which both start with $1$ and end in $00$. Thus, there are $5$ bits which we can vary, resulting in $n(Acap B)=2^5$.
The required answer is $2^7+2^6-2^5=128+64-32=160$.
answered Dec 17 '18 at 13:46
Shubham JohriShubham Johri
5,189718
$endgroup$
Recall that $$n(Acup B)=n(A)+n(B)-n(Acap B)$$
where $A$ is the set of $8$ length bit strings starting with $1,B$ is the set of $8$ length bit strings ending in $00$.
$n(A)=2^7$, because the remaining last $7$ bits can be $0,1$.
$n(B)=2^6$, because the remaining first $4$ bits can be $0,1$.
$n(Acap B)$ is the number of $8$ length bit strings which both start with $1$ and end in $00$. Thus, there are $5$ bits which we can vary, resulting in $n(Acap B)=2^5$.
The required answer is $2^7+2^6-2^5=128+64-32=160$.
answered Dec 17 '18 at 13:46
Shubham JohriShubham Johri
5,189718
answered Dec 17 '18 at 13:46
Shubham JohriShubham Johri
5,189718
answered Dec 17 '18 at 13:46
Shubham JohriShubham Johri
5,189718
answered Dec 17 '18 at 13:46
Shubham JohriShubham Johri
5,189718
5,189718
add a comment |
add a comment |
2
$begingroup$
Assuming you meant the non-exclusive or: There are $2^7$ that begin with $1$. There are $2^6$ that end in $00$. There are $2^5$ that BOTH begin with $1$ and end in $00$. Thus the answer is $2^7+2^6-2^5=160$. Assuming you meant the exclusive or: we then have to subtract $2^5$ again, now getting your $128$. So...did you mean the exclusive or? That should be specified.
$endgroup$
– lulu
Dec 17 '18 at 13:45
$begingroup$
@lulu I noticed that you put you answerS as comments. Why? (just curious)
$endgroup$
– adhg
Dec 18 '18 at 1:11
$begingroup$
Can you add how you got your answer of $128$? We'll be able to help you better if we can see how you got your answer.
$endgroup$
– Carl Schildkraut
Dec 18 '18 at 4:26
1
$begingroup$
@adhg That was hardly an answer...I was just pointing out that the question was ambiguous and that (what I imagine to be) the standard reading would yield a different answer but that a (somewhat) non-standard reading would give the "right" answer. I was asking the OP for clarification.
$endgroup$
– lulu
Dec 18 '18 at 12:00
$begingroup$
I think "Either" in this question signify 'Exclusive or' itself and there is no need to specify it.
$endgroup$
– Apoorv Garg
Dec 19 '18 at 8:07