wilNajbribe
Answered

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 :

toporc
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:
[tex]9\times8\times7\times6\times5\times4\times3\times2\times1=362,880\ ways [/tex]
meerkat18
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. 

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