GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 06 Aug 2020, 03:11

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Students in a certain class are about to participate in debate teams

Author Message
TAGS:

### Hide Tags

Intern
Joined: 11 Dec 2017
Posts: 1
Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

07 Jul 2020, 11:25
3
00:00

Difficulty:

95% (hard)

Question Stats:

25% (02:37) correct 75% (02:12) wrong based on 73 sessions

### HideShow timer Statistics

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
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 2062
Location: India
Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

07 Jul 2020, 14:32
Suppose we put the married couple in a team. Now the number of ways to choose the third member of the team = 7C1

Total possible combinations of such teams $$= \frac{7C1*6C3*3C3}{2! }= 70$$ [We need to divide by 2!, since the other 2 teams are similar]

tamarlvn 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
Intern
Joined: 13 Aug 2018
Posts: 1
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

07 Jul 2020, 18:06
Could you please elaborate why have you added 3C3 in this combination set?

Posted from my mobile device
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 2062
Location: India
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

07 Jul 2020, 18:20
1
After selecting the team that includes married couple, we left with 6 students.

Number of ways to select second team = Number of ways to select 3 students out of 6= 6C3

Now we left with 3 students.

Number of ways to select third team= 3C3 = 1

Shresta5 wrote:
Could you please elaborate why have you added 3C3 in this combination set?

Posted from my mobile device
Manager
Joined: 12 Apr 2018
Posts: 69
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

11 Jul 2020, 02:25
tamarlvn 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

Make two cases:
Case 1 - Team that have married couple
Case 2 - Teams without married couple

Case 1 * Case 2 -> Our three desired team

Case 1 -> Ways to select first two member of team -> Only one way (Married couple) -> Ways to select third team member out of 7 remaining member -> 7
Hence, number of ways to select team 1 -> 1*7 = 7

Case 2 ->
Number of ways to select second team or T2 of three member -> 6c3 = 20
Number of ways to select third team or T3 of three member -> 3c3 = 1

Since the order of selection does not matter in team selection, our cases will be reduced by half
Because first combination -> [ T2-(m1,m2,m3) T3-(m4,m5,m6)] is same as second combination -> [ T2-(m4,m5,m6) T3-(m1,m2,m3)]
Hence, possible combinations will be 20/2=10

Thus, total number of cases will be = Case 1 * Case 2
= 7*10
= 70
Manager
Status: BELIEVE IN YOURSELF
Joined: 06 Oct 2019
Posts: 98
Location: India
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

14 Jul 2020, 19:08
I'm getting 140.

Posted from my mobile device
Intern
Joined: 12 Apr 2020
Posts: 34
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

14 Jul 2020, 23:17
nick1816 wrote:
Suppose we put the married couple in a team. Now the number of ways to choose the third member of the team = 7C1

Total possible combinations of such teams $$= \frac{7C1*6C3*3C3}{2! }= 70$$ [We need to divide by 2!, since the other 2 teams are similar]

tamarlvn 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

Dear Sir,

Could you please elaborate logic for 2!. I went for 140.

DS Forum Moderator
Joined: 19 Oct 2018
Posts: 2062
Location: India
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

15 Jul 2020, 04:28
Sir nhi bhai chalega!

Since we can't differentiate between 2 groups in which we don't have married couple, we have to divide by 2!. Logic is pretty similar to the following question.

Find the number of arrangements of the letters of the word "OFF"?

rishit924 wrote:
nick1816 wrote:
Suppose we put the married couple in a team. Now the number of ways to choose the third member of the team = 7C1

Total possible combinations of such teams $$= \frac{7C1*6C3*3C3}{2! }= 70$$ [We need to divide by 2!, since the other 2 teams are similar]

tamarlvn 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

Dear Sir,

Could you please elaborate logic for 2!. I went for 140.

Manager
Joined: 12 Apr 2018
Posts: 69
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

15 Jul 2020, 05:45
Shresta5 wrote:
Could you please elaborate why have you added 3C3 in this combination set?

Posted from my mobile device

Think it in the way that you have now only three members left to be picked and you have to make the third team of three people.

Let's say you have three person A, B and C left and you have to make a team of three people comprising of A, B and C.

Now think of possible ways you can select the team from these people ->
Way 1 -> You select A first, then B and finally C -> Final result is ABC
Way 2 -> You select B first, then C and finally A -> Final result is BCA
Way 3 -> You select C first, then A and finally B -> Final result is CAB
Way 4 -> You select A first, then C and finally B -> Final result is ACB
Way 5 -> You select B first, then A and finally C -> Final result is BAC
Way 6 -> You select C first, then B and finally A -> Final result is CBA

But if you'll notice, you are getting only one actual combination and order does not matter as team consist of only those three people.

Thus, there is only one way to select such team and it can be represented as 3C3 -> Selection of 3 from a set of 3
Manager
Joined: 18 Feb 2020
Posts: 165
Location: India
Schools: ISB '21
GMAT 1: 660 Q50 V29
GPA: 3
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

15 Jul 2020, 21:30
1
The OA is 70. Option (C). The major confusion is between 140 and 70. I'll try explaining.

For the married couple to remain together, consider them as one entity. There are 7 other people in the class.
If anyone member gets along with the couple, those 3 can form a debate team.

So the number of ways in which a team can be formed containing the married couple = 7C1 = 7.

Now the remaining 6 people need to be arranged in groups of 3.
Therefore, selecting 3 people out of 6 = 6C3 = 20 ---- (A)

Let us consider a case. Members are named A - G and the couple are simply H and W.

Example case :

H, W, A are together --- Team 1 ( 7C1)
B, C, D are together --- Team 2 (6C3)
E, F, G are together --- Team 3

Now when B, C, and D are selected in 6C3; E, F, and G are automatically selected as well for Team 3. There would also be one other case where you'll select E, F, and G for Team 1. In that case, B, C, and D would form Team 3.

Carefully understand that both the cases are identical since the order of teams doesn't matter. That's why we divide by 2!.

Hence, (6C3*7C1)/2! = 70.

Option (C)
Manager
Joined: 18 Feb 2020
Posts: 165
Location: India
Schools: ISB '21
GMAT 1: 660 Q50 V29
GPA: 3
Re: Students in a certain class are about to participate in debate teams  [#permalink]

### Show Tags

15 Jul 2020, 21:33
rishit924 wrote:
nick1816 wrote:
Suppose we put the married couple in a team. Now the number of ways to choose the third member of the team = 7C1

Total possible combinations of such teams $$= \frac{7C1*6C3*3C3}{2! }= 70$$ [We need to divide by 2!, since the other 2 teams are similar]

tamarlvn 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

Dear Sir,

Could you please elaborate logic for 2!. I went for 140.

I hope you find the above explanation helpful!
Re: Students in a certain class are about to participate in debate teams   [#permalink] 15 Jul 2020, 21:33