There are 11 members in a family out of which there are 4 males and remaining females. The family has hired three cars for a trip to a zoo. The members are to seated in the cars in such a way that there are not more than four members in one car and there is at least one male in each car. How many different ways can the members travel?

Assume there are 3 cars,total 11 members so in 2 cars 4 members and in one car 3 members should be there.
So in first car,second car let us assume 4 members are there in each car atleast 1 should be male.
Total males-4 ,females-7
Car1 selections:4c1*7c3+4c2*7c2+4c3*7c1+4c4*7c0=295
Similarly in car2 also selections 295
Car3 selections:only 3 members out of that atleast one male
So 4c1*7c2+4c2*7c1+4c3*7c0=130
So total ways are 295+295+130=720.

Number of people in car can be (4, 4, 3) only.
Now, we have 4 males for 3 cars. So, it has to be (2, 1, 1).
Now, we get two cases: 2 men in a car of 3 and 2 men in a car of 4.

Case 1: Two men in a car of 3
These two men can be selected in 4C2 ways. Then, they can pick a car in 3 ways. Remaining two men can pick cars in 2*1 ways.
Now, we have 7 women and (1, 3, 3) psostions in car.
One lady can be chosen in 7 ways. Second car can have ladies in 6C3 ways while other car will have remaining ladies.

So, total = 4C2 * 3 * 2*1 * 7 * 6C3 ways = 5040 ways

Case 2: Two men in a car of 4
These two men can be selected in 4C2 ways and they can pick a car in 3 ways. Remaining men can pick cars in 2*1 ways.
Now, we have 7 women and (2, 3, 2) posistions in cars of capacity (4, 4, 3) resp.
So, first car can have women in 7C2 ways. Second car can have them in 5C3 ways and remaining two will go to third car.
One more point here that the cars which have 1 man each, have capacity of 4 and 3. So, the car with capacity of 4 can be chosen in 2 ways. Other will have capacity of 1.

So, total = 4C2 * 3 * 2*1 * 7C2 * 5C3 * 2*1 ways = 15120 ways

In all, we have (5040+15120) ways = 20160 ways

630 ways ?