Mix of permutation and combination a car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if John and Samantha must sit in the back seat and there is only one driver? the answer is 288, but i didn't get it >_< my solution I drew 5 box 1 box represent driver -> 5C1 2 box represent front-> 5C2 4 box represent back-> 6C4*2C2 which will lead to 5C1*5C2*6C4*2C2=750

The problem says only one person can drive. We can choose $2$ people from the remaining $4$ to join S and J in the back seat in $\binom{4}{2}$ ways. The $4$ people in the back seat can arrange themselves in $4!$ ways, and the $2$ non-drivers in front can arrange themselves in $2!$ ways, for a total of $\binom{4}{2}4!2!$.