How many ways could three people sit at a table with five seats in which two of the five seats will remain empty?
A. 8
B. 12
C. 60
D. 118
E. 120
The OA is C.
Ways in which 3 seats out 5 can be selected = 5C3
Ways in which 3 people can be arranged in 3 seats = 3!
Total ways of arrangement = 5C3*3!
= 5*4! / 2*6
= 10*6
= 60.
Has anyone another suggestion about how to solve this PS question? Thanks!
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How many ways could three people sit at a table
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BTGmoderatorLU
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Let the 3 people be A, B and C.BTGmoderatorLU wrote:How many ways could three people sit at a table with five seats in which two of the five seats will remain empty?
A. 8
B. 12
C. 60
D. 118
E. 120
Number of options for A = 5. (Any of the 5 seats.)
Number of options for B = 4. (Any of the 4 remaining seats.)
Number of options for C = 3. (Any of the 3 remaining seats.)
To combine these options, we multiply:
5*4*3 = 60.
The correct answer is C.
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BTGmoderatorLU wrote:How many ways could three people sit at a table with five seats in which two of the five seats will remain empty?
A. 8
B. 12
C. 60
D. 118
E. 120
The number of choices for a seat the first person could have is 5. Once he has sat down, the number of choices for a seat the second person could have is 4. Finally, the number of choices for a seat the third person, after the first two have sat down, could have is 3. Thus, the number of ways 3 people can sit a table of with five seats is:
5 x 4 x 3 = 60
Alternate Solution:
The number of seats to be occupied can be selected in 5C3 = 5!/(3!*2!) = (5x4)/2 = 10 ways.
Once the seats to be occupied are determined, there are 3! = 6 ways the three people can be arranged in those seats.
In total, there are 10 x 6 = 60 ways the three people can sit at a table of five seats.
Answer: C
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