eran_m wrote:Students in a certain class are about to participate in debate teams, which are composed of 3 students each. How many different combinations are possible, if there are 9 students in the class, and the only married couple in the class must be on the same team?
a. 10
b. 42
c. 70
d. 140
e. 1,680
[spoiler]
OA C.
[/spoiler]
Team that includes the married couple:
Number of options for the third member of this team = 7. (There are 7 people besides the married couple.)
Two remaining teams:
The remaining 6 people must be divided into two teams of 3.
Number of options for the first team = 6C3 = (6*5*4)/(3*2*1) = 20.
Number of options for the second team = 3C3 = (3*2*1)/(3*2*1) = 1.
To combine these options, we multiply:
20*1 = 20.
Since the ORDER of these two teams does not matter -- dividing the 6 remaining people into ABC-DEF is the same as dividing them into DEF-ABC -- the result above must be divided by the number of ways the two teams can be ARRANGED (2!):
20/2! = 10.
To combine the options above, we multiply:
7*10 = 70.
The correct answer is
C.
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