Thread: Probability...i Think nCR

Eight kids, 4 boys, 4 girls arrange themselves at random around a merry go round. What is the probabilty that they will be seated alternately.....?

i cant figure out how to do it...8! ways they could be arranged...but order matters (girls boys) but not between girls and between boys..so how do i do it?

Originally Posted by stones44

Eight kids, 4 boys, 4 girls arrange themselves at random around a merry go round. What is the probabilty that they will be seated alternately.....?

This is a circular arrangement problem. N individuals can be arranged in a circle in (N-1)! Ways.
So there are (7!) ways to arrange the children.
Now place the girls in every other set: (3!) ways to do that.
But now the seats on the ride are ordered, so there are (4!) ways to seat the boys.
What is the answer?

ok so its 4!3! / 7! thanks!....

new thread

a bridge hand which contains no card higher than a 9 is called a yarborough. what is the probability that one sucj hand will appear in a deal consisting of four hands. what is the probability that two will appear in adeal.

Originally Posted by bjhopper

new thread

a bridge hand which contains no card higher than a 9 is called a yarborough. what is the probability that one sucj hand will appear in a deal consisting of four hands. what is the probability that two will appear in adeal.

Hmm, be more specific ? u say cards, is it a pack of unbiased cards of play ?

bridge is played with a standard deck of 52 cards. i assume you know the suits and faces

Originally Posted by Plato

This is a circular arrangement problem. N individuals can be arranged in a circle in (N-1)! Ways.
So there are (7!) ways to arrange the children.
Now place the girls in every other set: (3!) ways to do that.
But now the seats on the ride are ordered, so there are (4!) ways to seat the boys.
What is the answer?

So N-1 is because there are N! - N ways in which they can be seated with the same order????