I don't suppose it would something like :
2C1 groups 2C1 b or girl 2n!/n! 2n!/n! all that would be the #ways you can get and equal # of boys and girls in each grouo: 2*2*(2n!/n!)^2
Then the total # ways to divide them would be
4nC2n * 2nC2n?
so the probability would be
4*[(2n)!/n!]^2 / 4nC2n
meh I'm just making this up kinda so XP
ok, i came up with an empirical solution, but i don't know if it is actually right.
Assuming n=1 (2 boys and 2 girls), then there will be 3 different pairs of groups :
b1 with g1 and b2 with g2
b1 with g2 and b2 with g1
b1 with b2 and g1 with g2
out of these 3 pairs of groups only 2 of them have the same number of boys and girls on each side, so:
2/3*100% = 66.67% ,...
that or 4n*(2n!)^2 / (4n!) .... :S
Well, in order to get the thrad going, I'll post the answer now.
There are 2n boys and 2n girls, so the total amount of people is 4n. Therefore there are 4nC2n possible combinations (C stands for combination). We must then consider the favorable combinations, i.e. in how many ways you can pick out n boys and n girls.
There are 2nCn ways of choosing n boys out of 2n. We also need exactly n girls, so the same probability applies there too. Therefore, there are (2nCn)^2 favourable combinations.
The probability can therefore be calculated by dividing the favorable combinations with the possible combinations, i.e. (2nCn)^2 / 4nC2n. After crunching the numbers, it should end up as ((2n)!)^4 / ((4n!) * (n!)^4).
This is the exact answer, but it's also possible to approximate the value of n! to simplify the formula. Anyway, whoever feels like it can post the next problem.
Alright, here's an easy one:
An object of mass m1 is n1 meters from the center of a spherical shell of thickness n2, with radius n3 from the center to the outer surface of the shell, where n3 > n1 (the object is within the shell).
The shell has a uniform of density of m0/(n0)^3
Assuming universal gravitation F=(G*mass1*mass2)/r^2, where G is some constant, what is the force experienced by the object?
dammit I hate physics.
dammit, I can't believe I tried thinking about this answer. I despiseeeeeeeeeeeeeeeeeee physics with a passionnnnnnnnnnnnnnnnnnn.
Though, I'm confused about why you're giving all the data.
F= ma, where a is just g.
Idk why you need all that extra information... especially if nothing is orbiting something...
i hate physics... i'm no good at it =(
^ It's because the G in question isn't the same one you'd experience on earth, isn't it?
Anyway prepare for some very shakey, highschool-level mechanics guys.
Mass of the shell = volume of 'anulus' multiplied by density : (4/3*π*n3^3 - 4/3*π*(n3-n2)^3) * m0/n0^3 let's call this β
so since I'm guessing this is taking place in space and the object in question hasn't just fallen to the bottom of the sphere, the only force acting on it would be gravitational attraction to the sphere. then F= (G*m1*β)/r^2 where r is the distance between the object and the part of the inner surface of the sphere closest to it?
F=(G*m1*β)/ ((n3-n2)-n1)^2 ?
No to either answer.
This really shouldn't be that hard though, and I'm pretty sure that this is a problem that you can google an answer for (haven't tried yet though).
m2 is the mass of the hollow sphere, and equals the volume times the density. The volume is obviously (4*pi*n3^3)/3 - (4*pi*(n3-n2)^3)/3, as jamie already stated, and the mass can easily be calculated from that. G is the gravitational constant, and m1 isn't any problem either. Therefore, the only open question is the value of r.
Originally Posted by adonai
I'm not a physicist, but the gravity in the centre of the spere should be 0 right? This implies that the value of r=0 in the centre of the sphere, as all other parts of the equation are positive constants. Therefore the force should be F=(G*m1*m2)/(n1^2), as the difference in r between the center and n1 obviously is n1.
This is really very simple, if you want to do it the hard the way it will involve quite a bit of calculus, but it's really not needed.
Just think of how gravity works for any solid spherical object.
I'll give the answer if anyone asks or don't get the next try.
^Don't give it before I give it a hand, please.