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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Step1: Lets take one person from 8. Now in how many ways can he be teamed up with another man? 7!!! Team-1 is formed. Step2: Lets take another person from remaining 6. In how many ways can he be teamed up with another man? 5!!! Team-2 is formed. Step3: Lets take another person from remaining 4. In how many ways can he be teamed up with another man? 3!!! Team-3 is formed. Step4: Last two can be teamed up in only 1 way.

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way. For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two are already chosen); For the third one in 3 ways (as 4 people are already chosen); For the fourth one there is only one left.

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

By multiplying all the combinations of different groups you are getting a figure for which order matters

For example if there are 8 people: john,mary,jane,rima,sandeep,huhu,bruce and ram

the groups (john,mary)(jane,rima)(sandeep,huhu)(bruce,ram) is the same as (bruce,ram)(john,mary)(sandeep,huhu)(jane,rima)

as mentioned earlier by multiplying the number of various combinations possible you get all the different ways the groups could be organised without differentiating that order does not patter. The way to discount this is to divide by the factorial of the number of items = 4!

Hi, I didn't know that we had to divide by 4!, I just knew that we had to divide by some number to make up for the repetitions. So I didn't know what to do with the 840 I kept on getting. So I left PnC and just tried another logic:

Here it is..

I took up a smaller group of 4 persons, whom I have to divide into 2 teams - A B C D

So the 1st team could have any one of AB, AC or AD - 3 and the other team in each of the cases has only one option - the other 2 persons. -- So it becomes 3 x 1 When I extended it to 6 persons into 2 teams, I got - 5 x 3 x 1

So I extended till 8 and got it 7 x 5 x 3 x 1 = 105

Could you please explain how you came up with each number: ((28)(15)(6)(1))/(4)(3)(2) = 105.

I am confused. Would be really helpful.

get 2 people from 8 = \(\frac{8!}{6!*2!}\)=28 get 2 people from 8-2 = \(\frac{6!}{4!*2!}\)=15 get 2 people from 8-4 = \(\frac{4!}{2!*2!}\)=6 get 2 people from 8-6 = \(\frac{2!}{0!*2!}\)=1

May be someone with a good understanding of permutations and combinations can explain this discrepency. Personally I never doubt the explanations given by AkamaiBrah but i dont know about this particular question. May be mods can help.

Step1: Lets take one person from 8. Now in how many ways can he be teamed up with another man? 7!!! Team-1 is formed. Step2: Lets take another person from remaining 6. In how many ways can he be teamed up with another man? 5!!! Team-2 is formed. Step3: Lets take another person from remaining 4. In how many ways can he be teamed up with another man? 3!!! Team-3 is formed. Step4: Last two can be teamed up in only 1 way.

So total ways = 7*5*3*1 = 105

For poor chaps like me who find the permutation, combination, probablity as a demon, your answers were simple to understand buddy....

+1 to u _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

I still have some trouble to figure out when to account for repetitions . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices. To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen? (OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy

I still have some trouble to figure out when to account for repetitions . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices. To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen? (OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy

Simplify the question in this way, you need to select a team of two from 8 = 8c2 = 28 now 6 are left and = 6c2 = 15 4c2 = 6 and finally only two people left =1

But you can select these in 4! ways. so final selection will be (8c2 * 6c2 * 4c2 * 2c2) / (4 !) = 105

I still have some trouble to figure out when to account for repetitions . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices. To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen? (OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy

Thank you so much

Elzi

I think the trick is not to get too technical, but to rely on logic. Here's how I did that (and feel free to correct my thinking)

-I have 8 people and I need to figure out how many DIFFERENT ways I can split these guys into 4 teams of 2. I'm making a chain of 4 teams. A chain is different only when there's a new team in there, not a different arrangement of the same 4 teams! -Let me call these teams T1, T2, T3, T4 -Clearly what the question is saying is that I don't care if I pick them the following ways, it's all the same to the author T2 T1 T3 T4 T2 T3 T1 T4 and 22 other varieties (i.e. a total of 4! or 24 ways I can arrange these 4 SPECIFIC teams) - ok, noted -Let's see how I can choose T1, T2, T3, T4 T1 - I can choose 8C2 ways T2 - I can choose 6C2 ways T3 - I can choose 4C2 ways T4 - I can choose 2C2 or 1 way (since well there are only 2 guys left)

The key is NOT to get confused on the usage of combinations and assume that since we're using combinations the order already doesnt matter. Sure in the last team say only Jack and Jill were left - we still only picked one team - i.e. we didn't distinguish between (Jack-Jill) or (Jill-Jack). This doesnt mean that we are done. We need to focus on what the question is asking us and we need to give them DISTINCT ways of PICKING 4 TEAMS of 2 (i.e. this is one further layer)

8C2 * 6C2 * 4C2 * 2C2 = total number of ways I can arrange Teams 1, 2, 3 and 4. But I don't care about how these teams are arranged. Therefore, I divide by 4!

It helps to try and understand what the author is asking rather than get stuck on what formula to use.

I'm also trying to solidify my understanding of when to adjust for arrangements etc. so this is work-in-progress!