suppose there are 8 people (A,B,C,D,E,F,G,H respectively).
in our class it was said that the probability that person A sitting before person B is 8!/2.
I'm little confused about how we got that answer.
any help would be appreciated!
in our class it was said that the probability that person A sitting before person B is 8!/2.
I'm little confused about how we got that answer.
any help would be appreciated!
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It's not the probability, it's the number of different seating arrangements. Obviously it can't be a probability because 8!/2 is a lot bigger than the highest possible probability which is 1.
There are 8 choices for the first seat, 7 for the second, 6 for the third and so on.
So there are 8! ways of seating the people.
In half of those, A is sitting before B.
So there are 8!/2 ways of sitting the people so that A is before B.
There are 8 choices for the first seat, 7 for the second, 6 for the third and so on.
So there are 8! ways of seating the people.
In half of those, A is sitting before B.
So there are 8!/2 ways of sitting the people so that A is before B.