Mixing
Given 8 People in a race including Alice and Bob. How many ranked lists of these people see Alice ahead of...
$begingroup$
This was an exam question on a paper I sat today.
My solution was something like this.
- Number of ways of arranging 8 unique things is 8!
- There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob.
- Therefore 8! / 2 is the number of combinations / ranked lists.
Is this correct? Thanks in advance.
combinatorics combinations
asked Jan 22 at 23:46
Jonathan BartlettJonathan Bartlett
1134
$endgroup$
add a comment |
$begingroup$
This was an exam question on a paper I sat today.
My solution was something like this.
- Number of ways of arranging 8 unique things is 8!
- There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob.
- Therefore 8! / 2 is the number of combinations / ranked lists.
Is this correct? Thanks in advance.
combinatorics combinations
asked Jan 22 at 23:46
Jonathan BartlettJonathan Bartlett
1134
$endgroup$
$begingroup$
Are ties between racers a possibility?
$endgroup$
– Daniel Schepler
Jan 22 at 23:59
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@DanielSchepler this wasn't something I thought of. I don't, however, believe that it was a possibility.
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:00
$begingroup$
Your answer is correct. But when you say "There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob." sounds like weird. You may say "We can separate the $8!$ exactly two distinct parts one of them Alice ahead of Bob and the other one Bob ahead of Alice. And the parts has same elements..."
$endgroup$
– S.S.Danyal
Jan 23 at 0:09
$begingroup$
@S.S.Danyal thank you
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:10
add a comment |
$begingroup$
This was an exam question on a paper I sat today.
My solution was something like this.
- Number of ways of arranging 8 unique things is 8!
- There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob.
- Therefore 8! / 2 is the number of combinations / ranked lists.
Is this correct? Thanks in advance.
combinatorics combinations
asked Jan 22 at 23:46
Jonathan BartlettJonathan Bartlett
1134
$endgroup$
This was an exam question on a paper I sat today.
My solution was something like this.
- Number of ways of arranging 8 unique things is 8!
- There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob.
- Therefore 8! / 2 is the number of combinations / ranked lists.
Is this correct? Thanks in advance.
combinatorics combinations
combinatorics combinations
asked Jan 22 at 23:46
Jonathan BartlettJonathan Bartlett
1134
asked Jan 22 at 23:46
Jonathan BartlettJonathan Bartlett
1134
asked Jan 22 at 23:46
Jonathan BartlettJonathan Bartlett
1134
asked Jan 22 at 23:46
Jonathan BartlettJonathan Bartlett
1134
asked Jan 22 at 23:46
Jonathan BartlettJonathan Bartlett
1134
1134
$begingroup$
Are ties between racers a possibility?
$endgroup$
– Daniel Schepler
Jan 22 at 23:59
$begingroup$
@DanielSchepler this wasn't something I thought of. I don't, however, believe that it was a possibility.
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:00
$begingroup$
Your answer is correct. But when you say "There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob." sounds like weird. You may say "We can separate the $8!$ exactly two distinct parts one of them Alice ahead of Bob and the other one Bob ahead of Alice. And the parts has same elements..."
$endgroup$
– S.S.Danyal
Jan 23 at 0:09
$begingroup$
@S.S.Danyal thank you
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:10
add a comment |
$begingroup$
Are ties between racers a possibility?
$endgroup$
– Daniel Schepler
Jan 22 at 23:59
$begingroup$
@DanielSchepler this wasn't something I thought of. I don't, however, believe that it was a possibility.
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:00
$begingroup$
Your answer is correct. But when you say "There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob." sounds like weird. You may say "We can separate the $8!$ exactly two distinct parts one of them Alice ahead of Bob and the other one Bob ahead of Alice. And the parts has same elements..."
$endgroup$
– S.S.Danyal
Jan 23 at 0:09
$begingroup$
@S.S.Danyal thank you
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:10
$begingroup$
Are ties between racers a possibility?
$endgroup$
– Daniel Schepler
Jan 22 at 23:59
$begingroup$
Are ties between racers a possibility?
$endgroup$
– Daniel Schepler
Jan 22 at 23:59
$begingroup$
@DanielSchepler this wasn't something I thought of. I don't, however, believe that it was a possibility.
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:00
$begingroup$
@DanielSchepler this wasn't something I thought of. I don't, however, believe that it was a possibility.
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:00
$begingroup$
Your answer is correct. But when you say "There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob." sounds like weird. You may say "We can separate the $8!$ exactly two distinct parts one of them Alice ahead of Bob and the other one Bob ahead of Alice. And the parts has same elements..."
$endgroup$
– S.S.Danyal
Jan 23 at 0:09
$begingroup$
Your answer is correct. But when you say "There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob." sounds like weird. You may say "We can separate the $8!$ exactly two distinct parts one of them Alice ahead of Bob and the other one Bob ahead of Alice. And the parts has same elements..."
$endgroup$
– S.S.Danyal
Jan 23 at 0:09
$begingroup$
@S.S.Danyal thank you
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:10
$begingroup$
@S.S.Danyal thank you
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:10
add a comment |
1 Answer
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$begingroup$
Here is the long answer with all the stuations.
Alice is first. Then put Bob to the one of the 7 rooms. And put other guys the other 6 rooms. Then we get $7.6!$
Alice is second. Then put Bob to the one of the 6 rooms. And put other guys the other 6 rooms. Then we get $6.6!$
Alice is third. Then put Bob to the one of the 5 rooms. And put other guys the other 6 rooms. Then we get $5.6!$
...
Alice is 6-th. Then put Bob to the one of the 2 rooms. And put other guys the other 6 rooms. Then we get $2.6!$.
Alice is 7-th. Then put Bob to the last room. And put other guys the other 6 rooms. Then we get $1.6!$.
Total sum $7.6!+6.6!+ldots+1.6!=6!(7+6+ldots+1)=6!.frac{7.8}{2}=frac{8!}{2}$.
answered Jan 23 at 0:19
S.S.DanyalS.S.Danyal
5666
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is the long answer with all the stuations.
Alice is first. Then put Bob to the one of the 7 rooms. And put other guys the other 6 rooms. Then we get $7.6!$
Alice is second. Then put Bob to the one of the 6 rooms. And put other guys the other 6 rooms. Then we get $6.6!$
Alice is third. Then put Bob to the one of the 5 rooms. And put other guys the other 6 rooms. Then we get $5.6!$
...
Alice is 6-th. Then put Bob to the one of the 2 rooms. And put other guys the other 6 rooms. Then we get $2.6!$.
Alice is 7-th. Then put Bob to the last room. And put other guys the other 6 rooms. Then we get $1.6!$.
Total sum $7.6!+6.6!+ldots+1.6!=6!(7+6+ldots+1)=6!.frac{7.8}{2}=frac{8!}{2}$.
answered Jan 23 at 0:19
S.S.DanyalS.S.Danyal
5666
$endgroup$
add a comment |
$begingroup$
Here is the long answer with all the stuations.
Alice is first. Then put Bob to the one of the 7 rooms. And put other guys the other 6 rooms. Then we get $7.6!$
Alice is second. Then put Bob to the one of the 6 rooms. And put other guys the other 6 rooms. Then we get $6.6!$
Alice is third. Then put Bob to the one of the 5 rooms. And put other guys the other 6 rooms. Then we get $5.6!$
...
Alice is 6-th. Then put Bob to the one of the 2 rooms. And put other guys the other 6 rooms. Then we get $2.6!$.
Alice is 7-th. Then put Bob to the last room. And put other guys the other 6 rooms. Then we get $1.6!$.
Total sum $7.6!+6.6!+ldots+1.6!=6!(7+6+ldots+1)=6!.frac{7.8}{2}=frac{8!}{2}$.
answered Jan 23 at 0:19
S.S.DanyalS.S.Danyal
5666
$endgroup$
add a comment |
$begingroup$
Here is the long answer with all the stuations.
Alice is first. Then put Bob to the one of the 7 rooms. And put other guys the other 6 rooms. Then we get $7.6!$
Alice is second. Then put Bob to the one of the 6 rooms. And put other guys the other 6 rooms. Then we get $6.6!$
Alice is third. Then put Bob to the one of the 5 rooms. And put other guys the other 6 rooms. Then we get $5.6!$
...
Alice is 6-th. Then put Bob to the one of the 2 rooms. And put other guys the other 6 rooms. Then we get $2.6!$.
Alice is 7-th. Then put Bob to the last room. And put other guys the other 6 rooms. Then we get $1.6!$.
Total sum $7.6!+6.6!+ldots+1.6!=6!(7+6+ldots+1)=6!.frac{7.8}{2}=frac{8!}{2}$.
answered Jan 23 at 0:19
S.S.DanyalS.S.Danyal
5666
$endgroup$
Here is the long answer with all the stuations.
Alice is first. Then put Bob to the one of the 7 rooms. And put other guys the other 6 rooms. Then we get $7.6!$
Alice is second. Then put Bob to the one of the 6 rooms. And put other guys the other 6 rooms. Then we get $6.6!$
Alice is third. Then put Bob to the one of the 5 rooms. And put other guys the other 6 rooms. Then we get $5.6!$
...
Alice is 6-th. Then put Bob to the one of the 2 rooms. And put other guys the other 6 rooms. Then we get $2.6!$.
Alice is 7-th. Then put Bob to the last room. And put other guys the other 6 rooms. Then we get $1.6!$.
Total sum $7.6!+6.6!+ldots+1.6!=6!(7+6+ldots+1)=6!.frac{7.8}{2}=frac{8!}{2}$.
answered Jan 23 at 0:19
S.S.DanyalS.S.Danyal
5666
answered Jan 23 at 0:19
S.S.DanyalS.S.Danyal
5666
answered Jan 23 at 0:19
S.S.DanyalS.S.Danyal
5666
answered Jan 23 at 0:19
S.S.DanyalS.S.Danyal
5666
5666
add a comment |
add a comment |
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$begingroup$
Are ties between racers a possibility?
$endgroup$
– Daniel Schepler
Jan 22 at 23:59
$begingroup$
@DanielSchepler this wasn't something I thought of. I don't, however, believe that it was a possibility.
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:00
$begingroup$
Your answer is correct. But when you say "There is a 50% chance Bob beats Alice and a 50% chance Alice beats Bob." sounds like weird. You may say "We can separate the $8!$ exactly two distinct parts one of them Alice ahead of Bob and the other one Bob ahead of Alice. And the parts has same elements..."
$endgroup$
– S.S.Danyal
Jan 23 at 0:09
$begingroup$
@S.S.Danyal thank you
$endgroup$
– Jonathan Bartlett
Jan 23 at 0:10