Given 8 People in a race including Alice and Bob. How many ranked lists of these people see Alice ahead of Bob?

Question

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.

Answer 1

Here is the long answer with all the stuations.

1. Alice is first. Then put Bob to the one of the 7 rooms. And put other guys the other 6 rooms. Then we get $7.6!$

2. Alice is second. Then put Bob to the one of the 6 rooms. And put other guys the other 6 rooms. Then we get $6.6!$

3. Alice is third. Then put Bob to the one of the 5 rooms. And put other guys the other 6 rooms. Then we get $5.6!$

...

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 $2.6!$.

7. 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 $7.6!+6.6!+\ldots+1.6!=6!(7+6+\ldots+1)=6!.\frac{7.8}{2}=\frac{8!}{2}$.