Cybermusings wrote:A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
28
32
48
60
12

Please provide alternative solutions

OA coming in after a while

...well since i dont know what ur solution is .. i cant comment on wether the solution i used is a alternate one or not ....

... nywayz i dont know if there culd be any other solution ... other than the one taking into account evry possible case ...

suppose the father is driving he car ..

1st case .. the mother is sitting in the front seat next to the father .. in the back the brother will have to sit in the middle and the girls next to him this canm be done in 2! ways ..

2nd case .. the brother sits in the front seat .. in this case the mother will sit in the middle and the girls can be seated in 2! ways ..

3rd case ... one of the girls sits in the front seat (this can be done in 2 ways ).. in this case the people in the back can be seated in 3! ways ... so total no. of seating arrangement 3!*2 = 12

So adding all the cases we have 2+2+12 = 16 ways .. all these cases can also be repeated when the mother is driving .. so total no. of possiblities = 16*2 = 32 ways ..

I'm with Gabriel on this. Can't figure out a simpler way than to
construct all possible (well only 3) combinations.

I'm getting the same 2 * ( 2 + 2 + 2*3!) = 32 as well