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Question Stats:
29% (01:52) correct
70% (01:21) wrong based on 2 sessions
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? (A) 90 (B) 105 (C) 168 (D) 420 (E) 2520 Source: GMAT Club Tests - hardest GMAT questions pl explain the ans particularly usage of 4!.
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Here is another conceptual way: 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
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vaivish1723 wrote: pl explain the ans particularly usage of 4!. 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! I hope that makes sense ((28)(15)(6)(1))/(4)(3)(2) = 105.
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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 
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vaivish1723 wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? (A) 90 (B) 105 (C) 168 (D) 420 (E) 2520 Source: GMAT Club Tests - hardest GMAT questions pl explain the ans particularly usage of 4!. , 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. So we have 7*5*3*1=105 Answer: B. You can check similar problems: probability-88685.html?hilit=different%20items%20divided%20equallyprobability-85993.html?highlight=divide+groupscombination-55369.html#p690842sub-committee-86346.html?highlight=divide+groupsThere is also direct formula for this: 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}Hope it helps.
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In!/Out! 8!/(4!2!2!2!2!) = 7 * 5 * 3 = 105
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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.
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ileannaconst wrote: 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
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\frac{1}{4!}*[\frac{8*7}{2!}*\frac{6*5}{2!}*\frac{4*3}{2!}*\frac{2*1}{2!}] = 105\frac{1}{4!} because order of groups does not matter.
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vaibhavtripathi wrote: Here is another conceptual way:
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
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Elzi wrote: Hi, 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 This concept is explained here: probability-88685.html?hilit=different%20items%20divided%20equallyprobability-85993.html?highlight=divide+groupscombination-55369.html#p690842sub-committee-86346.html?highlight=divide+groups
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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.
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if we dont go by the brilliant method suggested by srijaidev and instead go with conventional combinatorics as suggested by nightwing: WHAT SHOULD BE THE WORDING OF THE QUESTION SO THAT WE DO NOT NEED TO DIVIDE THE TOTAL NUMBER OF COMBINATIONS BY 4! ? I am posting this query so that I can understand the difference in the 2 question types if faced by either on G-day. 
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Hi, 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
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What's the probability that combinatorics appear on the exam?
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Thanks a lot Bunuel!
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Another option is to use the formula (found it in one of the threads):
(m*n)!/[(n!)^m]*m!
m - number of people
n - number of groups
8!/[(2)^4]*4! = 105
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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
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