How many ways could three people sit at a table

[BTGmoderatorLU]

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!

[GMATGuruNY]

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

Let the 3 people be A, B and C.
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.

[regor60]

Or find permutation of 5 "different" things, 5!, and then adjust for two that are the same, 2! =

5!/2! = 5x4x3 = 60

[Scott@TargetTestPrep]

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