Anurag591 wrote:
Thank You AnupamGMAT760 for catering my doubt but there is a constrain question that is :
"No Family Member Will Use Same Car" and there are 5 different CARS for 6 people.
So you have to allocate 1 A brand car to George thus 5!. This is similar to questions, where 3 will come at last position or first position in a 4 digit number and find possible combinations. Please help me understand where I am getting it wrong
Hi Anurag
Consider this:
All family members have a fixed position lets say son gets spot 1, father spot 2 and so on. Now to allocate different cars to them we have to keep these members fixed at their places while we arrange the cars in different spots. Say car A gets spot 1 so it will go to the son since we assigned him that spot and that spot remains fixed. Now there are many ways to approach this:
You can arrange these cars in different spots without constraints in 6!/2! ways. 2! is used since the two A cars are identical and all arrangements involving first A car will be the same as arrangements involving the second A car.
Now factor the constraints, assign Car A to the son so your first spot gets fixed. Now 5 different cars can be arranged in 5 different spots in 5! ways.
We want cases where the son does not get Car A hence (6!/2!) - 5!= 360-120= 240
Alternatively, you also could have used the direct method:
Son A has 4 options to choose from and others still in their respective slots can be assigned a car in 5!/2! ways (5 spots- 5 arrangements- 2 identical)
4*60=240