A contractor wishes to build 9 houses, each of different design. In how many ways can heplace these homes on a street if 6 lots are on one side of the street and 3 lots are on theopposite side?
The Correct Answer and Explanation is :
To solve this problem, let’s break it down into steps.
The contractor has 9 houses, each with a different design. He needs to place them on a street with 6 lots on one side and 3 lots on the opposite side. The goal is to determine how many ways the contractor can arrange the houses on the 9 available lots.
Step 1: Select the 3 houses for the opposite side of the street
There are 9 houses in total. The first task is to choose 3 houses to place on the side with 3 lots. The number of ways to choose 3 houses from 9 is a combination, and it can be calculated using the combination formula:
[
C(n, k) = \frac{n!}{k!(n-k)!}
]
where (n) is the total number of items (houses), and (k) is the number of items to choose. In this case, we calculate:
[
C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84
]
So, there are 84 ways to choose 3 houses to place on the side with 3 lots.
Step 2: Arrange the 3 selected houses on the opposite side
Once the 3 houses are chosen for the 3 lots, these houses can be arranged in any order. The number of ways to arrange 3 houses on 3 lots is simply the number of permutations of 3 houses, which is:
[
3! = 3 \times 2 \times 1 = 6
]
Step 3: Arrange the remaining 6 houses on the other side
After selecting and arranging the 3 houses for the opposite side, there are 6 remaining houses. These 6 houses can be arranged on the 6 lots on the other side of the street. The number of ways to arrange these 6 houses is:
[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
]
Step 4: Multiply all the possibilities
Now, we can multiply the possibilities for each step:
[
84 \times 6 \times 720 = 362880
]
Thus, the total number of ways the contractor can place the houses on the street is 362,880.
Explanation of the Process:
This problem involves both combinations and permutations. First, we use combinations to select which houses go on the side with fewer lots (3 houses), and then we use permutations to arrange the selected houses on the available lots. For the remaining 6 houses, the number of possible arrangements is calculated using permutations. Multiplying the results from each step gives us the final number of possible arrangements.