Question

A contractor wishes to build 9 houses, each different in design. In how many ways can he place these houses on a street if 6 lots are on one side of the street and 3 lots are on the opposite side?

Answer 1

There are 9 choices of lot for the first house, 8 choices of lot for the second house, 7 choices for the third house. Continuing this reasoning the total number of ways the houses can be placed is:

`9*8*7*6*5*4*3*2*1=362,880\ ways`

Answer 2

So base on your problem that ask how many ways can he place these houses on a street if 6 lots are on one side of the street and 3 lots on the opposite side and the best answer based on my calculation is 9C6 = 9!/(6!*3!). I hope you understand my answer.