Cannot figure this

[eran_m]

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]

[jkaustubh]

I think that there is something missing in the question, as ther is no information about the number of the married couples.

Please tell me, if I am wrong or not!!

[GMATGuruNY]

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.