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?

The houses can be placed in 362,880 ways.

Step-by-step explanation:

The 9 houses are each in different design.

The each lot can place any of the 9 houses.

Similarly, the process gets repeated until the last house is placed on a lot.

From the above steps, it can be determined that :

The number of ways to place the 9 houses in 9 lots = 9!

⇒ 9×8×7×6×5×4×3×2×1

⇒ 362880 ways.

Therefore, the houses can be placed in 362880 ways.

for the points

step-by-step explanation:

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8