Question

Question #3cd19

Answer 1

210 ways

Explanation:

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