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!/5!
(E) 10!
[spoiler]OA=D[/spoiler]
I am confused. Could anyone give me a detailed explanation here? Please. I'd be thankful.
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A lawn care company has five employees, and there are ten
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Gmat_mission
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The posted problem has been transcribed from the original source -- the book GMAT for Dummies -- incorrectly.
In the book, the prompt reads as follows:
Number of houses that could be assigned to the second employee = 9. (Any of the 9 remaining houses.)
Number of houses that could be assigned to the third employee = 8. (Any of the 8 remaining houses.)
Number of houses that could be assigned to the fourth employee = 7. (Any of the 7 remaining houses.)
Number of houses that could be assigned to the fifth employee = 6. (Any of the 6 remaining houses.)
To combine these options, we multiply:
10*9*8*7*6.
The product above is equivalent to D:
10!/5! = 10*9*8*7*6.
The correct answer is D.
In the book, the prompt reads as follows:
Number of houses that could be assigned to the first employee = 10. (Any of the 10 houses.)Gmat_mission wrote: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 provides service for just one home?
(A) 50
(B) 2!/1!
(C) 120
(D) 10!/5!
(E) 10!
Number of houses that could be assigned to the second employee = 9. (Any of the 9 remaining houses.)
Number of houses that could be assigned to the third employee = 8. (Any of the 8 remaining houses.)
Number of houses that could be assigned to the fourth employee = 7. (Any of the 7 remaining houses.)
Number of houses that could be assigned to the fifth employee = 6. (Any of the 6 remaining houses.)
To combine these options, we multiply:
10*9*8*7*6.
The product above is equivalent to D:
10!/5! = 10*9*8*7*6.
The correct answer is D.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
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regor60
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Number of ways to select 5 houses from 10: 10!/5!*5!Gmat_mission wrote: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!/5!
(E) 10!
[spoiler]OA=D[/spoiler]
I am confused. Could anyone give me a detailed explanation here? Please. I'd be thankful.
Number of ways a group of 5 houses can be distributed among 5 employees: 5!
Total number of ways: (10!/5!*5!)*5! = [spoiler]10!/5!, D[/spoiler]
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Jeff@TargetTestPrep
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Number of ways the houses could be assigned to the first employee = 10Gmat_mission wrote: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 provides service for just one home?
(A) 50
(B) 2!/1!
(C) 120
(D) 10!/5!
(E) 10!
Number of ways the houses could be assigned to the second employee = 9
Number of ways the houses could be assigned to the third employee =8
Number of ways the houses could be assigned to the fourth employee = 7
Number of ways the houses that could be assigned to the fifth employee = 6
So the total number of ways is
10 x 9 x 8 x 7 x 6 = 10!/5!
Answer: D
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