Question #3cd19

1 Answer
Oct 20, 2016

210 ways


You start by placing Alice, there are 3 ways. then there are are 2 ways of placing Jack. So we have 6 ways of placing Alice and Jack.
In the car with Alice, there are #¹⁰C₃=(10!)/(7!*3!)=(10*9*8)/(1*2*3)=120# ways for the first car.

For the second car, you may or may not have Jack.
If Jack is in the second car, there are #⁷C₃=(7!)/(4!*3!)=(7*6*5)/(1*2*3)=35# ways for the second car.

For the third car, there is #1# way.
If Jack is not in the third car, then we have #⁷C₄=(7!)/(3!*4!)=35# ways for the second car and 1 way for the third car.

Putting it all together #=6*(35)=210# ways