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D
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Difficulty:
55%
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Question Stats:
61% (02:02) correct
39%
(02:10)
wrong
based on 176
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A lawn care company has five employees, and there are ten houses that need care on a given day. How many different ways can the company assign the five employees to work at the different houses on that day if each employee works at two houses?
(A) 50
(B) 2!/1!
(C) 120
(D) 10!/32
(E) 10!
(A) 50
(B) 2!/1!
(C) 120
(D) 10!/32
(E) 10!
24 May 2017, 05:33
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rohan2345
Nice question!
Take the task of assigning two houses to each employee and break it into stages.
Stage 1: Select 2 houses for employee #1
Since the order in which we select the two houses does not matter, we can use combinations.
We can select 2 houses from 10 houses 10C2 ways [= (10)(9)/(2) ways]
So, we can complete stage 1 in (10)(9)/(2) ways
NOTE: Given the answer choices, I'm not going to evaluate (10)(9)/(2). You'll see why shortly.
Stage 2: Select 2 houses for employee #2
There are 8 houses left to choose from.
So we can complete stage 2 in 8C2 ways (= (8)(7)/(2) ways)
Stage 3: Select 2 houses for employee #3
There are 6 houses left to choose from.
We can complete stage 3 in 6C2 ways (= (6)(5)/(2) ways)
Stage 4: Select 2 houses for employee #4
We can complete stage 4 in 4C2 ways (= (4)(3)/(2) ways)
Stage 5: Select 2 houses for employee #5
We can complete stage 5 in 2C2 ways (= (2)(1)/(2) ways)
By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus assign all 10 houses to 5 employees) in [(10)(9)/(2)][(8)(7)/(2)][(6)(5)/(2)][(4)(3)/(2)][(2)(1)/(2)] ways
Evaluate to get: 10!/(2^5)
Answer: . . . hmmm, not among the answer choices.
I Googled the question and found out that it's a question from GMAT For Dummies. They say the answer is D, but give no rationale, other than to plug 10 and 5 into the permutations formula. However, this calculation would be for a scenario in which each worker works at ONLY ONE house (not 2, as in the question)
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.
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[you-tube]https://www.youtube.com/watch?v=05a0A7vjG8I[/you-tube
13 Jun 2018, 16:51
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rohan2345
Number of ways the houses could be assigned to the first employee = 10C2. (Any 2 of the 10 houses.)
Number of ways the houses could be assigned to the second employee = 8C2. (Any 2 of the 8 remaining houses.)
Number of ways the houses could be assigned to the third employee = 6C2. (Any 2 of the 6 remaining houses.)
Number of ways the houses could be assigned to the fourth employee = 4C2. (Any 2 of the 4 remaining houses.)
Number of ways the houses that could be assigned to the fifth employee = 2C2. (Any 2 of the 2 remaining houses.)
So the total number of ways is
10C2 x 8C2 x 6C2 x 4C2 x 2C2
(10 x 9)/2! x (8 x 7)/2! x (6 x 5)/2! x (4 x 3)/2! x (2 x 1)/2!
10!/(2!)^5
10!/2^5
10!/32
(Answer is not there.)
