combinatorics problems on number of ways a student can answer a a paper of two sections .
$begingroup$
A question paper is split into two parts – part $A$ and part $B$. Part $A$ contains $5$ questions and part $B$ has $4$ questions. Each question in part $A$ has an alternative. A student has to attempt at least one question from each part. Find the number of ways in which the student can attempt the question paper.
This question tells that part $A$ has $5$ question with an alternative. does that mean that we have $10$ or $5$ question to choose from .
for part $A$ should I choose the number of ways as $(binom{10}1 + binom{10}2 + binom{10}3 + binom{10}4 + binom{10}5)$ or $(binom51 + binom52 + binom53 + binom54 + binom55)$.
combinatorics
edited Jan 11 at 14:57
Namaste
1
asked Jan 11 at 14:35
lamdeblamdeb
11
$endgroup$
|
show 1 more comment
$begingroup$
A question paper is split into two parts – part $A$ and part $B$. Part $A$ contains $5$ questions and part $B$ has $4$ questions. Each question in part $A$ has an alternative. A student has to attempt at least one question from each part. Find the number of ways in which the student can attempt the question paper.
This question tells that part $A$ has $5$ question with an alternative. does that mean that we have $10$ or $5$ question to choose from .
for part $A$ should I choose the number of ways as $(binom{10}1 + binom{10}2 + binom{10}3 + binom{10}4 + binom{10}5)$ or $(binom51 + binom52 + binom53 + binom54 + binom55)$.
combinatorics
edited Jan 11 at 14:57
Namaste
1
asked Jan 11 at 14:35
lamdeblamdeb
11
$endgroup$
1
$begingroup$
I think the instructions are unclear. Consider question A1 and its "alternative" A1a. I think it means a student can attempt one or the other, but not both. So I would count $3^5-1$ ways of completing part A.
$endgroup$
– Gerry Myerson
Jan 11 at 14:52
$begingroup$
The way I read this there are five questions in part A. For each such question, there are two versions. Thus, if you pick a question from part A, you have two choices of which version you answer. However, the way it reads, I don't think you can answer both versions of the same question and count them as two of the five questions you can answer.
$endgroup$
– N. F. Taussig
Jan 12 at 12:54
$begingroup$
Any thoughts, lamdeb?
$endgroup$
– Gerry Myerson
Jan 13 at 1:39
$begingroup$
Earth to lamdeb, come in, please.
$endgroup$
– Gerry Myerson
Jan 14 at 3:43
$begingroup$
I am sorry ,I did not see your comment as i had gone out .. I haven't understood why you chose the 3^5 instead of 2^5 and why the -1 . the question already tells that the student must pick at least 1 question . so there is no way where he cannot pick no question.
$endgroup$
– lamdeb
Jan 14 at 11:26
|
show 1 more comment
$begingroup$
A question paper is split into two parts – part $A$ and part $B$. Part $A$ contains $5$ questions and part $B$ has $4$ questions. Each question in part $A$ has an alternative. A student has to attempt at least one question from each part. Find the number of ways in which the student can attempt the question paper.
This question tells that part $A$ has $5$ question with an alternative. does that mean that we have $10$ or $5$ question to choose from .
for part $A$ should I choose the number of ways as $(binom{10}1 + binom{10}2 + binom{10}3 + binom{10}4 + binom{10}5)$ or $(binom51 + binom52 + binom53 + binom54 + binom55)$.
combinatorics
edited Jan 11 at 14:57
Namaste
1
asked Jan 11 at 14:35
lamdeblamdeb
11
$endgroup$
A question paper is split into two parts – part $A$ and part $B$. Part $A$ contains $5$ questions and part $B$ has $4$ questions. Each question in part $A$ has an alternative. A student has to attempt at least one question from each part. Find the number of ways in which the student can attempt the question paper.
This question tells that part $A$ has $5$ question with an alternative. does that mean that we have $10$ or $5$ question to choose from .
for part $A$ should I choose the number of ways as $(binom{10}1 + binom{10}2 + binom{10}3 + binom{10}4 + binom{10}5)$ or $(binom51 + binom52 + binom53 + binom54 + binom55)$.
combinatorics
combinatorics
edited Jan 11 at 14:57
Namaste
1
asked Jan 11 at 14:35
lamdeblamdeb
11
edited Jan 11 at 14:57
Namaste
1
asked Jan 11 at 14:35
lamdeblamdeb
11
edited Jan 11 at 14:57
Namaste
1
edited Jan 11 at 14:57
Namaste
1
edited Jan 11 at 14:57
Namaste
1
1
asked Jan 11 at 14:35
lamdeblamdeb
11
asked Jan 11 at 14:35
lamdeblamdeb
11
asked Jan 11 at 14:35
lamdeblamdeb
11
11
1
$begingroup$
I think the instructions are unclear. Consider question A1 and its "alternative" A1a. I think it means a student can attempt one or the other, but not both. So I would count $3^5-1$ ways of completing part A.
$endgroup$
– Gerry Myerson
Jan 11 at 14:52
$begingroup$
The way I read this there are five questions in part A. For each such question, there are two versions. Thus, if you pick a question from part A, you have two choices of which version you answer. However, the way it reads, I don't think you can answer both versions of the same question and count them as two of the five questions you can answer.
$endgroup$
– N. F. Taussig
Jan 12 at 12:54
$begingroup$
Any thoughts, lamdeb?
$endgroup$
– Gerry Myerson
Jan 13 at 1:39
$begingroup$
Earth to lamdeb, come in, please.
$endgroup$
– Gerry Myerson
Jan 14 at 3:43
$begingroup$
I am sorry ,I did not see your comment as i had gone out .. I haven't understood why you chose the 3^5 instead of 2^5 and why the -1 . the question already tells that the student must pick at least 1 question . so there is no way where he cannot pick no question.
$endgroup$
– lamdeb
Jan 14 at 11:26
|
show 1 more comment
1
$begingroup$
I think the instructions are unclear. Consider question A1 and its "alternative" A1a. I think it means a student can attempt one or the other, but not both. So I would count $3^5-1$ ways of completing part A.
$endgroup$
– Gerry Myerson
Jan 11 at 14:52
$begingroup$
The way I read this there are five questions in part A. For each such question, there are two versions. Thus, if you pick a question from part A, you have two choices of which version you answer. However, the way it reads, I don't think you can answer both versions of the same question and count them as two of the five questions you can answer.
$endgroup$
– N. F. Taussig
Jan 12 at 12:54
$begingroup$
Any thoughts, lamdeb?
$endgroup$
– Gerry Myerson
Jan 13 at 1:39
$begingroup$
Earth to lamdeb, come in, please.
$endgroup$
– Gerry Myerson
Jan 14 at 3:43
$begingroup$
I am sorry ,I did not see your comment as i had gone out .. I haven't understood why you chose the 3^5 instead of 2^5 and why the -1 . the question already tells that the student must pick at least 1 question . so there is no way where he cannot pick no question.
$endgroup$
– lamdeb
Jan 14 at 11:26
1
1
$begingroup$
I think the instructions are unclear. Consider question A1 and its "alternative" A1a. I think it means a student can attempt one or the other, but not both. So I would count $3^5-1$ ways of completing part A.
$endgroup$
– Gerry Myerson
Jan 11 at 14:52
$begingroup$
I think the instructions are unclear. Consider question A1 and its "alternative" A1a. I think it means a student can attempt one or the other, but not both. So I would count $3^5-1$ ways of completing part A.
$endgroup$
– Gerry Myerson
Jan 11 at 14:52
$begingroup$
The way I read this there are five questions in part A. For each such question, there are two versions. Thus, if you pick a question from part A, you have two choices of which version you answer. However, the way it reads, I don't think you can answer both versions of the same question and count them as two of the five questions you can answer.
$endgroup$
– N. F. Taussig
Jan 12 at 12:54
$begingroup$
The way I read this there are five questions in part A. For each such question, there are two versions. Thus, if you pick a question from part A, you have two choices of which version you answer. However, the way it reads, I don't think you can answer both versions of the same question and count them as two of the five questions you can answer.
$endgroup$
– N. F. Taussig
Jan 12 at 12:54
$begingroup$
Any thoughts, lamdeb?
$endgroup$
– Gerry Myerson
Jan 13 at 1:39
$begingroup$
Any thoughts, lamdeb?
$endgroup$
– Gerry Myerson
Jan 13 at 1:39
$begingroup$
Earth to lamdeb, come in, please.
$endgroup$
– Gerry Myerson
Jan 14 at 3:43
$begingroup$
Earth to lamdeb, come in, please.
$endgroup$
– Gerry Myerson
Jan 14 at 3:43
$begingroup$
I am sorry ,I did not see your comment as i had gone out .. I haven't understood why you chose the 3^5 instead of 2^5 and why the -1 . the question already tells that the student must pick at least 1 question . so there is no way where he cannot pick no question.
$endgroup$
– lamdeb
Jan 14 at 11:26
$begingroup$
I am sorry ,I did not see your comment as i had gone out .. I haven't understood why you chose the 3^5 instead of 2^5 and why the -1 . the question already tells that the student must pick at least 1 question . so there is no way where he cannot pick no question.
$endgroup$
– lamdeb
Jan 14 at 11:26
|
show 1 more comment
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1
$begingroup$
I think the instructions are unclear. Consider question A1 and its "alternative" A1a. I think it means a student can attempt one or the other, but not both. So I would count $3^5-1$ ways of completing part A.
$endgroup$
– Gerry Myerson
Jan 11 at 14:52
$begingroup$
The way I read this there are five questions in part A. For each such question, there are two versions. Thus, if you pick a question from part A, you have two choices of which version you answer. However, the way it reads, I don't think you can answer both versions of the same question and count them as two of the five questions you can answer.
$endgroup$
– N. F. Taussig
Jan 12 at 12:54
$begingroup$
Any thoughts, lamdeb?
$endgroup$
– Gerry Myerson
Jan 13 at 1:39
$begingroup$
Earth to lamdeb, come in, please.
$endgroup$
– Gerry Myerson
Jan 14 at 3:43
$begingroup$
I am sorry ,I did not see your comment as i had gone out .. I haven't understood why you chose the 3^5 instead of 2^5 and why the -1 . the question already tells that the student must pick at least 1 question . so there is no way where he cannot pick no question.
$endgroup$
– lamdeb
Jan 14 at 11:26