Question #3cd19

1 Answer
Narad T. · Solace M.
Oct 20, 2016

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

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