Probability that there will be 2 passengers in each car 12 persons get on a train that has six cars;each passenger may select with equal probability each of the cars.If $P_1$ be the probability that there will be 1 car without passenger,1 with one passenger,2 with two passengers each and the remaining 2 with three and four passengers respectively and $P_2$ be the probability that there will be 2 passengers in each car,then find the ratio $\frac{P_1}{P_2}$.
I found $P_2$ as $\frac{12!}{(2!)^6}$ but i could not find $P_1$.Please help me.
 A: What you have computed as $P_2$ is actually $N_2$, the # of ways.  
However, it won't matter if we are just to compute the ratio $\dfrac{P_1}{P_2}$
Making the same assumption as you, that the six cars are distinct,
$\dfrac{P_1}{P_2} = \dfrac{\dbinom{12}{0,1,2,2,3,4}}{\dbinom{12}{2,2,2,2,2,2,}}= \dfrac{(2!)^4}{3!4!}$
PS.
Since the cars are taken to be distinct, the numerator needs to be multiplied by $\dfrac{6!}{2!}$ to take care of permutations of the pattern of distribution. You will then get the answer of $40$.
A: A warning: This calculation is correct, but much more verbose than it should have been.  I tackled the problem logically and wound up with the correct answer—40.  This is a pure record of how I approached the problem.
The sample space here is $6^{12}$, because each of 12 passengers can choose among 6 distinct cars.
The Number of Ways (the numerator of $P_2$) is fairly simple: $\dbinom{12}2\dbinom{10}2\dbinom{8}2\dbinom{6}2\dbinom{4}2\dbinom{2}2$
Just choose 2 people for the first car, 2 for the second car, and so on, until you run out of people.
$P_1$ is a little harder because you have to allow for different possible orderings of the cars, whereas that is taken into account in the above formula for $N_2$.  So before we get to $P_1$, I'll show an alternate approach for calculating $N_2$.
In this approach, first I will count the number of ways to split 12 people into 6 groups of two (not counting the sequence of the groups, or the sequence of people in the groups—just the partitioning).  Then we multiply this by the number of ways to place 6 groups into 6 cars, which is obviously $6!$.
Partitioning (if that is the proper name) is a little bit tricky; I haven't had to do it before so I figured it out myself.  I'm sure there's a better notation for this, but you can simply "choose" each partition you want out of however many people you have left, and divide by the product of factorials of the counts of the number of groups of the same size.
That sounds unintelligible even to me, so let me give an example:  If you want to count the number of ways to split 18 people into 3 groups of 4 people and 1 group of 6 people, you can calculate it with:
$\frac{\dbinom{18}{4}\dbinom{14}{4}\dbinom{10}{4}\dbinom{6}{6}}{3!}$
The 3! in the denominator comes about because you've chosen 3 groups of 4 people, and you will have counted each separate partition 3! times.  (For completeness you could add a 1! in the denominator because of the single group of 6 people, but I chose to omit that constant factor.)
Likewise, if you want to count the ways of partitioning 35 people into 7,5,3,5,5,7,3, you can get it by:
$\frac{\dbinom{35}{7}\dbinom{28}{5}\dbinom{23}{3}\dbinom{20}{5}\dbinom{15}{5}\dbinom{10}{7}\dbinom{3}{3}}{(2!)(3!)(2!)}$
Because the counts of same size groups are 2,3,2—2 groups of 7, 3 groups of 5, 2 groups of 3.
(There MUST be a better notation, but this calculation method works and I'm going to use it.)  (NOTE: When I got to canceling, I saw why the simpler formula of $\dbinom{35}{3,3,5,5,5,7,7}$ actually works.  The above is still correct, it just includes more than it needs to.)
Back to $N_2$.  This new method gives the same answer I got before, divided by the factorials of the counts of same size groups (and there is just one size of group, and there are 6 of them, so that's 6! in the denominator), and then multiplied by the number of ways to distribute the 6 groups among 6 cars...which is 6! again.  These cancel out nicely, giving the answer already reached for $N_2$.
Now on to $N_1$.  Using my method as explained above, we get:
$\frac{\dbinom{12}{0}\dbinom{12}{1}\dbinom{11}{2}\dbinom{9}{2}\dbinom{7}{3}\dbinom{4}{4}}{(1!)(1!)(2!)(1!)(1!)}\times6!$
We can ignore the sample space for purpose of calculating $\frac{P_1}{P_2}$, so putting this all above $N_2$ and canceling, we get:
$\frac{2!2!2!6!}{3!4!} = 40$
(Now that I've worked through it myself, I understand what true blue anil's notation means—and where he went wrong, including an extra 2! instead of a 6! in the numerator, and how simple the calculation I outlined above actually is.  I tackled this with no formal training in Combinatorics, though, so I will leave it as-is—it's the record of a logical attack on the problem.)  ;)
@trueblueanil: I think you mixed up double counting the partitions of the two groups of two (which is compensated for by dividing by 2!) and the fact of 6 distinct cars.  The 6 distinct cars lead to an extra 6! in the numerator, not just a 2! from two groups same size.
