Calculating Probability: \binom{8}{2} \binom{6}{2} \binom{4}{2}\binom{2}{2}

[DottZakapa]

Homework Statement

Eight friends check in together at the reception of an hotel. There are 4 double rooms available, one at the 1st, one at the 2nd, one at the 3rd and one at the 7th floor.
If any possible assignment is equally likely, which is the probability that Alfred and John share the room at the 7th floor?

Relevant Equations

probability

My reasoning is :
first i evaluate in how many manners i can choose 2 guys from the group of 8.
$\binom{8}{2} \binom{6}{2} \binom{4}{2}\binom{2}{2}$

then i consider the probability of choosing the seventh floor is 1/4.
But now I don't now how to proceed, supposing up to here i am correct. any help?

 

[PeroK]

Homework Statement:: Eight friends check in together at the reception of an hotel. There are 4 double rooms available, one at the 1st, one at the 2nd, one at the 3rd and one at the 7th floor.
If any possible assignment is equally likely, which is the probability that Alfred and John share the room at the 7th floor?
Relevant Equations:: probability

My reasoning is :
first i evaluate in how many manners i can choose 2 guys from the group of 8.
$\binom{8}{2} \binom{6}{2} \binom{4}{2}\binom{2}{2}$

then i consider the probability of choosing the seventh floor is 1/4.
But now I don't now how to proceed, supposing up to here i am correct. any help?

What's the probability that Alfred is on the seventh floor?

 

[etotheipi]

Another approach would be to fix Alfred and John on the 7th floor, and determine the number of ways to put the other 6 people into the 3 remaining rooms. Then you can divide by the number of ways to put all 8 people into the 4 rooms.

 

[DottZakapa]

Another approach would be to fix Alfred and John on the 7th floor, and determine the number of ways to put the other 6 people into the 3 remaining rooms. Then you can divide by the number of ways to put all 8 people into the 4 rooms.

The number of ways to put the other 6 people into the 3 remaining rooms:

$\binom{6}{2} \binom{4}{2}\binom{2}{2}$
so
$\frac {\binom{6}{2} \binom{4}{2}\binom{2}{2}} {\binom{8}{2} \binom{6}{2} \binom{4}{2}\binom{2}{2}}$

that is:

$\frac {1}{\binom{8}{2}}$ = $\frac {1}{28}$

but could you explain why?

 

[PeroK]

but could you explain why?

You mean explain why that's the right answer?

 

[DottZakapa]

You mean explain why that's the right answer?

yes please

 

[PeroK]

yes please

The counting idea is:

a) You count all the different ways to do something: $N$.

b) You count all the ways to do something that meet your criteria: $n$.

Then, assuming all these ways are equally likely, the probability your criteria are met is $\frac n N$.

 

[DottZakapa]

a ok. thanks