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?

Question

Rhett Allison

Mathematics

29 June, 07:59

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?

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Answers (1)

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Victor Mann · Answer 1

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.

The 1st lot can place anyone house of all the 9 houses. The 2nd lot can place one of remaining 8 houses. The 3rd lot can place one of remaining 7 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.