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?

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?

directrix · Answer

The two sides of the street are not relevant to the problem.

There are 9 slots to fill with 9 differently designed houses.

Using the Fundamental Theorem of Counting,

there are 9 ways to select a house for the first slot, 8 ways to select a house for the second slot. Continuing, there will be 1 way to select a house for the 9th slot.

9! = 9 * 8 *7 * 6 * 5 * 4 * 3 * 2 * 1.

Multiply that out and see what the answer is. @Athaul

directrix · Answer

More practice problems here: http://mysite.dlsu.edu.ph/faculty/fillonea/documents/engstat%5Cex_probability.pdf

anonymous · Answer

there is another way to solve it.
select 6 houses(for one side) = 9C6
ways to arrange 6 houses = 6!
select 3 leftover houses = 3C3 =1
ways to arrange them = 3!
total number = 9C6*6!*3! = 9!

directrix · Answer

@Harindu   9 C 6 is not equal to 6!. I think you have the combination and permutation formulas mixed up.  

And, it is incorrect to state that 6! * 3! = 9! .

anonymous · Answer

lol 9C6 = 9!/(6!*3!)