I have two groups of 4 players each (8 players in total). Each player to play with other 7 players (in groups of 4) on 5 different days. Minimum one player will play with other plyer over the 5 days is 2, and the maximum is 3 days.
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In how many different ways can a group of 8 people be divide [#permalink]
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Question Stats: Hide Show timer StatisticsIn how many different ways can a group of 8 people be divided into 4 teams of 2 people each?A, 90B. 105C. 168D. 420 E. 2520
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Re: combination (groups and that stuff...) [#permalink]
The answer to the question is 105, not 2520. I posted a solution on another forum, which I'll paste here:________Think of this question: A group of eight tennis players will be divided into four teams of two. One team will play in the Olympics, one in Wimbledon, one in the Davis Cup and one in the US Open. In how many different ways can the teams be selected? Here, the order of the teams themselves clearly matters. If we choose {A,B} to go to the Olympics, and {C,D} to go to Wimbledon, that's clearly different from sending {C,D} to the Olympics and {A,B} to Wimbledon. The answer to this question is exactly the answer you give above:-you have 8C2 choices for the Olympics team;-you have 6C2 choices for the Wimbledon team;-you have 4C2 choices for the Davis Cup team;-you have 2C2 (one) choice for the US Open team.Multiply these to get the answer: 8C2*6C2*4C2*2C2 = (8*7/2)(6*5/2)(4*3/2)(2*1/2) = 2520.Note that the question I've just asked above is different from the question in the original post. In this question:A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people? the order of the teams does not matter. If we choose, say, these teams: {A,B}, {C,D}, {E,F}, {G,H}that's exactly the same set of teams as these:{C,D}, {A,B}, {G,H}, {E,F}Because the order of the teams themselves does not matter, we must divide by 4! = 24, the number of different orders we can put the four teams in, because all 24 different orders are in fact the same set of teams. So the answer is 2520/4! = 105. _________________
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Re: ways to divide?? [#permalink]
bibha wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520 \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), 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.So we have 7*5*3*1=105Answer: B.You can check similar problems:probability-88685.html?hilit=different%20items%20divided%20equally probability-85993.html?highlight=divide+groups combination-55369.html#p690842 sub-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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Re: In how many different ways can a group of 8 people be [#permalink] Another way to think about this:How many ways can you arrange the following:T1 T1 T2 T2 T3 T3 T4 T4That would be: 8!/(2!*2!*2!*2!)Then also recall that we don't care about the differences between the teams, therefore 8!/(2!*2!*2!*2!*4!) = 105
Intern Joined: 23 Aug 2012 Posts: 8
Re: ways to divide?? [#permalink]
Bunuel wrote: bibha wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520 \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), 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.So we have 7*5*3*1=105Answer: B.There 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. I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)?
Math Expert Joined: 02 Sep 2009 Posts: 87080
Re: ways to divide?? [#permalink]
dandarth1 wrote: Bunuel wrote: bibha wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each? 90 105 168 420 2520 \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), 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.So we have 7*5*3*1=105Answer: B.There 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. I understand how to get the answer, but I'm wondering why we assume that the order of the teams doesn't matter. The question just asks how many ways you can group them. When should we assume something matters or doesn't matter when the question doesn't specify? And also, just to clarify, when we do 8!/2!2!2!2!4!, we are also not worrying about the order within each team either right i.e. (a,b)=(b,a)? The teams are not numbered/labeled (we don't have team #1, #2, ...), the teams are not assigned to something (for example to tournaments), ... So, the order of the teams doesn't matter.Please check the links in my previous post for similar problems.Hope it helps. _________________
Manager Joined: 07 Feb 2011 Posts: 78
Re: In how many different ways can a group of 8 people be [#permalink] Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand? I am messing this up conceptually and want to correct this mistake
GMAT Expert Joined: 16 Oct 2010 Posts: 13246 Location: Pune, India
In how many different ways can a group of 8 people be divide [#permalink]
manimgoindowndown wrote: Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand? I am messing this up conceptually and want to correct this mistake It discusses two different questions:1. Distributing 12 different chocolates equally among 4 boys (similar to splitting 8 people in 4 distinct teams with 2 people each - your question)2. Distributing 12 different chocolates equally in 4 stacks (similar to splitting 8 people in 4 teams of 2 people each - the original question)See if grouping makes sense thereafter. _________________
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Originally posted by KarishmaB on 03 Mar 2013, 23:02.
Manager Joined: 07 Feb 2011 Posts: 78
Re: In how many different ways can a group of 8 people be [#permalink]
VeritasPrepKarishma wrote: manimgoindowndown wrote: Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand? I am messing this up conceptually and want to correct this mistake Check out this post: http://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html?/2011/11 ... ke-groups/ It discusses two different questions:1. Distributing 12 different chocolates equally among 4 boys (similar to splitting 8 people in 4 distinct teams with 2 people each - your question)2. Distributing 12 different chocolates equally in 4 stacks (similar to splitting 8 people in 4 teams of 2 people each - the original question)See if grouping makes sense thereafter. That example and wording was extremely confusing and frustrating. I am still trying to see what language prompted the difference in permutation vs combination.Maybe my brain has been fried this week (lots of pracatice, one full length CAT), but the big thing in this problem is you DO NOT account for order TWICEthe 4! is for the teamand the 2! is within every single combination for each pair
GMAT Expert Joined: 16 Oct 2010 Posts: 13246 Location: Pune, India
Re: In how many different ways can a group of 8 people be [#permalink]
manimgoindowndown wrote: That example and wording was extremely confusing and frustrating. I am still trying to see what language prompted the difference in permutation vs combination.Maybe my brain has been fried this week (lots of pracatice, one full length CAT), but the big thing in this problem is you DO NOT account for order TWICEthe 4! is for the team and the 2! is within every single combination for each pair Yes, because in this question, the groups are not distinct. You have to split 8 people in 4 groups.You can split them like this: (A, B), (C, D), (E, F), (G, H)or like this: (G, H), (A, B), (C, D), (E, F)they are the same split. They are not assigned to group1, group2, group3, group4. Say the total number of ways we get = NNow, if we change the question and say that we have 8 people and we need to divide them into 4 teams: Team 1, Team 2, Team 3 and Team 4Then, one split is this: Team 1 = (A, B); Team 2 = (C, D), Team 3 = (E, F), Team 4 = (G, H)and another split is: Team 1 = (G, H), Team 2 = (A, B); Team 3 = (C, D), Team 4 = (E, F)These two cases were the same in our original question but if the teams/groups are distinct, the two cases are distinct. Now, the total number of ways = N*4! _________________
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Re: In how many different ways can a group of 8 people be [#permalink]
bibha 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 Ways of choosing 2 out od a group of n people = nC2i.e. Ways of choosing first team of 2 out of 8 = 8C2i.e. Ways of choosing Second team of 2 out of remaining 6 = 6C2i.e. Ways of choosing Third team of 2 out of remaining 4 = 4C2Remaining 2 will form Forth teamTotal Ways of Choosing Teams = 8C2 * 6C2 * 4C2 * 1BUT Since the first team may come on second places and second may come on third etc. i.e.e arrangement among teams are included here which we NEED to excludei.e. Total Ways of Choosing Teams = 8C2 * 6C2 * 4C2 * 1 / 4! = 28*15*6/24 = 105Answer: option B _________________
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Re: In how many different ways can a group of 8 people be [#permalink]
bibha 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 The FIRST pair can be chosen in 8C2 waysThe SECOND pair can be chosen in 6C2 ways as 6 members are left after choosing 1st pairThe THIRD pair can be chosen in 4C2 ways as 4 members are left after choosing 1st two pair and so on...Therefore total ways of splitting 8 members in 6 pairs = (8C2 * 6C2 * 4C2 * 2C2) / 4! The entire expression is divided by 4! because the arrangements of pair have been accounted for, which need to be excluded. Arrangement for example Same pair which came at first place can be chosen at third place or forth places also while we are choosing different pairs at different places Required answer = (28 * 15 * 6 * 1)/24 = 105Answer: option C _________________
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Re: In how many different ways can a group of 8 people be [#permalink]
manalq8 wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?A. 90B. 105C. 168D. 420 E. 2520 Let's select the 4 groups of 2 individuals each8C2 * 6C2 * 4C2 * 2C2But wait... There is some problem...Let 8 individuals areA B C D E F G HCase 1: selected 4 groups of 2 are respectivelyA&B C&D E&F G&Hi.e. A&B us first group, C&D is second group etcCase 2: selected 4 groups of 2 are respectivelyC&D A&B G&H E&F respectivelySo case 1 and 2 are the same except that order of the groups has been accounted forThe arrangements of 4 groups can be done in 4! Which needs to be excluded from calculationHence divide the result by 4!So final result = 8C2 * 6C2 * 4C2 * 2C2 / 4! = 105Answer Option BHope this helps!!! _________________
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In how many different ways can a group of 8 people be divide [#permalink]
noboru wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?A, 90B. 105C. 168D. 420 E. 2520 Number of ways can a group of 8 people be divided into 4 teams of 2 people each = \(^8C_2*^6C_2*^4C_2*^2C_2/4!\) = 28*15*6*1 /4! = 105IMO B _________________
Kinshook Chaturvedi
Originally posted by Kinshook on 22 Aug 2019, 03:15.
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Re: In how many different ways can a group of 8 people be divide [#permalink]
noboru wrote: In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?A, 90B. 105C. 168D. 420 E. 2520 Responding to a pm:Quote: What if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be:8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values. I know from your post that if all four teams had different number of people we would not unarrange. But if some repeat? Yes, consider various scenarios:8 people and 2 teams (one of 3 people and the other of 5 people)You just pick 3 of the 8 people for the 3 people team. Rest everyone will be in the 5 people team. No un-arranging required.8 people and 2 teams (one of 4 people and the other of 4 people)You pick 4 people for the first team (say A, B, C, D). The other 4 (E, F, G, H) belong to the other team of 4 people. So you calculate this as 8C4. Now within this 8C4 will lie the case in which you picked (E, F, G, H) for the first team and ( A, B, C, D) will belong to the other team. But note that this case is exactly the same as before. 2 teams one (A, B, C, D) and other (E, F, G, H). So you need to un-arrange here.Similarly, take your case - 8 people - 3 teams of 1 individual in each and 1 team let's say consisting of 5 peopleWe select 5 people out of 8 for the 5 people team in 8C5 ways. Rest of the 3 people play individually and we need to create no teams so nothing to be done. Since we are not doing any selection for a team, we are not inadvertently arranging and hence un-arranging is not required. Instead say we have 8 people - and we make 3 teams, one with 4 people and two teams with 2 people eachWe select 4 people out of 8 for the 4 people team in 8C4 ways. Then we select 2 people for the first two people team in 4C2 ways and the leftover 2 people are for the second two people team. But again, as discussed above, there will same cases counted twice here so we will need to un-arrange by dividing by 2. _________________
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Re: In how many different ways can a group of 8 people be divide [#permalink] - (1) Captain from the first team has 7 options (excluding the captain). - (2) Two people are gone from (1), and we need to exclude the captain from the second team, giving us 6-2-1 = 5. Thus, 5 options left for the second captain.- (3) Four people are gone, Excluding the captain from the third team (8-4-1) we have 3 options left for the last captain. - (4) 7 * 5 * 3 =105.
Re: In how many different ways can a group of 8 people be divide [#permalink] 04 Oct 2022, 22:54 |
How many ways can you split 8 players into two teams of 4 players each?
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