Probability with restriction
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I have a question here which goes:
8 speakers are to speak one after another in a conference. Dr Cooper’s lecture is related to Dr Hofstadter’s and should not precede it. How many schedules could be arranged?
The solution is 8!/2 = 20160 which I understand how it is derived through the formula of permutation of non-distinct objects.
However, my first initial solution is 6! * 7P2 = 30240
Reasoning being, The 6 people that do not have any restriction I will arrange them first with order considered. Thereafter I slot the two other (Cooper and Hofstadter) in the 7 slots within the 6 people. I dont get which test case did I double counted in using this formula.
Please advice.
probability permutations
asked Nov 20 at 10:00
userName
124
add a comment |
up vote
0
down vote
favorite
I have a question here which goes:
8 speakers are to speak one after another in a conference. Dr Cooper’s lecture is related to Dr Hofstadter’s and should not precede it. How many schedules could be arranged?
The solution is 8!/2 = 20160 which I understand how it is derived through the formula of permutation of non-distinct objects.
However, my first initial solution is 6! * 7P2 = 30240
Reasoning being, The 6 people that do not have any restriction I will arrange them first with order considered. Thereafter I slot the two other (Cooper and Hofstadter) in the 7 slots within the 6 people. I dont get which test case did I double counted in using this formula.
Please advice.
probability permutations
asked Nov 20 at 10:00
userName
124
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a question here which goes:
8 speakers are to speak one after another in a conference. Dr Cooper’s lecture is related to Dr Hofstadter’s and should not precede it. How many schedules could be arranged?
The solution is 8!/2 = 20160 which I understand how it is derived through the formula of permutation of non-distinct objects.
However, my first initial solution is 6! * 7P2 = 30240
Reasoning being, The 6 people that do not have any restriction I will arrange them first with order considered. Thereafter I slot the two other (Cooper and Hofstadter) in the 7 slots within the 6 people. I dont get which test case did I double counted in using this formula.
Please advice.
probability permutations
asked Nov 20 at 10:00
userName
124
I have a question here which goes:
8 speakers are to speak one after another in a conference. Dr Cooper’s lecture is related to Dr Hofstadter’s and should not precede it. How many schedules could be arranged?
The solution is 8!/2 = 20160 which I understand how it is derived through the formula of permutation of non-distinct objects.
However, my first initial solution is 6! * 7P2 = 30240
Reasoning being, The 6 people that do not have any restriction I will arrange them first with order considered. Thereafter I slot the two other (Cooper and Hofstadter) in the 7 slots within the 6 people. I dont get which test case did I double counted in using this formula.
Please advice.
probability permutations
probability permutations
asked Nov 20 at 10:00
userName
124
asked Nov 20 at 10:00
userName
124
asked Nov 20 at 10:00
userName
124
asked Nov 20 at 10:00
userName
124
asked Nov 20 at 10:00
userName
124
124
add a comment |
add a comment |
1 Answer
1
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votes
up vote
2
down vote
You used $7P2$, that means Dr C and Dr H can come in any order but Dr C's should come after Dr H's lecture. You should use $7c2$.
Also, you're missing cases like HC123456 and 652HC134
answered Nov 20 at 10:26
Anvit
1,510419
In particular, there are $binom{7}{2}6!$ arrangements in which Dr. Hofstadter and Dr. Cooper are separated by at least one other speaker and $binom{7}{1}6!$ arrangements in which Dr. Cooper speaks immediately after Dr. Hofstadter.
– N. F. Taussig
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You used $7P2$, that means Dr C and Dr H can come in any order but Dr C's should come after Dr H's lecture. You should use $7c2$.
Also, you're missing cases like HC123456 and 652HC134
answered Nov 20 at 10:26
Anvit
1,510419
In particular, there are $binom{7}{2}6!$ arrangements in which Dr. Hofstadter and Dr. Cooper are separated by at least one other speaker and $binom{7}{1}6!$ arrangements in which Dr. Cooper speaks immediately after Dr. Hofstadter.
– N. F. Taussig
2 days ago
add a comment |
up vote
2
down vote
You used $7P2$, that means Dr C and Dr H can come in any order but Dr C's should come after Dr H's lecture. You should use $7c2$.
Also, you're missing cases like HC123456 and 652HC134
answered Nov 20 at 10:26
Anvit
1,510419
In particular, there are $binom{7}{2}6!$ arrangements in which Dr. Hofstadter and Dr. Cooper are separated by at least one other speaker and $binom{7}{1}6!$ arrangements in which Dr. Cooper speaks immediately after Dr. Hofstadter.
– N. F. Taussig
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
You used $7P2$, that means Dr C and Dr H can come in any order but Dr C's should come after Dr H's lecture. You should use $7c2$.
Also, you're missing cases like HC123456 and 652HC134
answered Nov 20 at 10:26
Anvit
1,510419
You used $7P2$, that means Dr C and Dr H can come in any order but Dr C's should come after Dr H's lecture. You should use $7c2$.
Also, you're missing cases like HC123456 and 652HC134
answered Nov 20 at 10:26
Anvit
1,510419
answered Nov 20 at 10:26
Anvit
1,510419
answered Nov 20 at 10:26
Anvit
1,510419
answered Nov 20 at 10:26
Anvit
1,510419
1,510419
In particular, there are $binom{7}{2}6!$ arrangements in which Dr. Hofstadter and Dr. Cooper are separated by at least one other speaker and $binom{7}{1}6!$ arrangements in which Dr. Cooper speaks immediately after Dr. Hofstadter.
– N. F. Taussig
2 days ago
add a comment |
In particular, there are $binom{7}{2}6!$ arrangements in which Dr. Hofstadter and Dr. Cooper are separated by at least one other speaker and $binom{7}{1}6!$ arrangements in which Dr. Cooper speaks immediately after Dr. Hofstadter.
– N. F. Taussig
2 days ago
In particular, there are $binom{7}{2}6!$ arrangements in which Dr. Hofstadter and Dr. Cooper are separated by at least one other speaker and $binom{7}{1}6!$ arrangements in which Dr. Cooper speaks immediately after Dr. Hofstadter.
– N. F. Taussig
2 days ago
In particular, there are $binom{7}{2}6!$ arrangements in which Dr. Hofstadter and Dr. Cooper are separated by at least one other speaker and $binom{7}{1}6!$ arrangements in which Dr. Cooper speaks immediately after Dr. Hofstadter.
– N. F. Taussig
2 days ago
add a comment |
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