The manager of a certain advertising agency must divide the group of 6

Question

The manager of a certain advertising agency must divide the group of 6 employees into a pair of teams with 3 employees each. One team will be the lead team and the other will be the backup team. How many different lead teams are possible?

(A) 120
(B) 60
(C) 40
(D) 20
(E) 10

(A) 120
(B) 60
(C) 40
(D) 20
(E) 10

FahedForero · Accepted Answer

That is correct if the order matters, but here you are just choosing a group so you are repeating terms within the group.

Your logic is missing the elimination of repeated elements.

So, 6*5*4 must be divided in 3! to get the number right

120/3! = 120/6 = 20

Bunuel · Answer

6C3 gives the number of all 3-member teams possible from 6, so that would be the number of lead teams possible.

Answer: D.

For similar questions check  DISTRIBUTING ITEMS/PEOPLE/NUMBERS... (QUESTION COLLECTION)

Hope it helps.

Kinshook · Answer

shreyaaaansh

Given: The manager of a certain advertising agency must divide the group of 6 employees into a pair of teams with 3 employees each. One team will be the lead team and the other will be the backup team.

Asked: How many different lead teams are possible?

Number of different lead teams = 6C3 = 20

IMO D

Kartike123 · Answer

6*5*4===120
1st position can be filled by 6 
2nd can be filled by 5 and 
3rd can be filled by 4 persons

Posted from my mobile device

Fish181 · Answer

I got 10 because the problem states that one team will be lead and one backup. So wouldn't that mean 1/2 of the teams would be lead and 1/2 would be backup therefore 10?

Bunuel · Answer

No. 6C3 gives the number of all possible 3-member teams from a pool of 6. For each case, one specific 3-member team out of 20 will be designated as the lead team, while the remaining 3 members will form the backup team.

Hope this helps.

ashutosh_73 · Answer

Bunuel  Below was my approach. I got the answer, but is it correct?

No of ways to make 2 groups = 6!/(3!x3!x2!) = 10
Within two groups, possibilities for a lead team = 2
Total possibilities = 10 x 2 = 20

Bunuel · Answer

Yes, that's correct - you are first finding the number of ways to divide 6 people into two groups and then multiplying by 2 since either one group can be the leading team. However, there is a much easier way of doing this as shown above.

MoPouyan · Answer

Step 1: Choose 3 employees for the lead team

The number of ways to select 3 employees out of 6 is given by the combination formula: 6C3

6C3=    = 20

So, there are 20 ways to choose 3 employees for the lead team.

Step 2: Assign the remaining 3 employees to the backup team

Once the lead team is chosen, the remaining 3 employees automatically form the backup team. Thus, there are no additional combinations to calculate for the backup team. 3C3=1

20 X 1=20 (We multiply because it is stated that both the lead team AND the backup team are considered.)

viswanath1 · Answer

Why dont we do 20*2 as there are 2c1 ways of selected which is the lead team

Bunuel · Answer

6C3 gives the number of all possible 3-member teams from a pool of 6. For each case, one specific 3-member team out of 20 will be designated as the lead team, while the remaining 3 members will form the backup team.

Hope this helps.

egmat · Answer

This is a classic combinations problem that trips up many students. Let me walk you through the key insight that makes this question much clearer.

The Critical Insight:

The moment you choose 3 people for the lead team, the remaining 3 people automatically become the backup team. You're not making two separate choices - you're really just asking "In how many ways can I select 3 people from 6 for the lead team?"

Let's think about this step by step:

Step 1: Understand what we're really selecting 
When the problem says "divide 6 employees into teams of 3 each," it sounds complicated. But notice that once you pick Alice, Bob, and Charlie for the lead team, Diana, Eve, and Frank must be the backup team. There's no separate choice to make.

Step 2: Recognize this as a combination problem 
Since we're selecting 3 people from 6, and the order within the team doesn't matter (Alice-Bob-Charlie is the same lead team as Bob-Alice-Charlie), this is a combinations problem, not permutations.

Step 3: Apply the combination formula 
We need  C(6,3) = \frac{6!}{3! 	imes 3!}

Let's calculate:  C(6,3) = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = \frac{120}{6} = 20

Step 4: Verify our logic 
Each selection of 3 people creates exactly one unique lead team and one unique backup team. Since we're only counting lead teams (not backup teams), our answer is 20.

Answer: (D) 20

The key breakthrough here is recognizing that you're not dividing into two separate groups - you're selecting one group, and the other is automatically determined. This insight applies to many similar "division" problems on the GMAT.

You can check out the  comprehensive solution on Neuron by e-GMAT  to master the systematic framework for all team selection problems and learn about the common trap answers that students fall into. You can also explore detailed solutions for  similar official questions  to build consistent accuracy across all combination problems.

Alilee · Answer

The question is asking you to choose lead teams ONLY. 
This is a combination Q, NOT a permutation Q.

(C) 40
 This is wrong because 6C3 * 2C1 is saying # of ways to choose the lead team "and" number of ways to be chosen as the lead team group. <- the question only asked for # of ways to choose lead group <- there is double couting here.  
(D) 20
 6C3 = # of ways to choose the lead teams, question only asks for # of ways to choose team A. nothing additional

sumkend · Answer

6C3 gives the number how many ways the first group could be made. it's 20

Usernamevisible · Answer

I multiple by 2 after 6c3 to assign lead team and backup team - failing to understand why is it wrong. What is the point of then saying lead and backup team. ?

Bunuel · Answer

Please check here: https://gmatclub.com/forum/the-manager- ... l#p3625721 Hope it helps.

Adit_ · Answer

If there was no demarcation of which is lead team and backup team then we would have to divide the result by 2 to give us 10.
Like when we have no of ways to seperate 6 people into two teams it gives us 10 but 6 choosing just 3 people out of 6 would be 20. A very interesting concept!

Adit_ · Answer

Since they have mentioned the lead and backup team it is 20. If there was no mention you have to divide the result by 2! = 20/2 = 10.

The manager of a certain advertising agency must divide the group of 6

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The manager of a certain advertising agency must divide the group of 6

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