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Re M2416 [#permalink]
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16 Sep 2014, 01:21
Official Solution:How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter) A. 8 B. 10 C. 16 D. 20 E. 24 \(\frac{C^3_6*C^3_3}{2!}=10\), we are dividing by \(2!\) as there are 2 groups and the order of the groups does not matter. For example consider the following 6 units: \(A\), \(B\), \(C\), \(D\), \(E\) and \(F\). Now, one of the groups that \(C^3_6\) gives is \(\{ABC\}\) and in this case the second group would be \(\{DEF\}\), so we have two groups \(\{ABC\}\) and \(\{DEF\}\). But \(C^3_6\) also gives \(\{DEF\}\) group and in this case the second group would be \(\{ABC\}\), so we have the same two groups: \(\{ABC\}\) and \(\{DEF\}\). Therefore to get rid of such duplications we should divide \(C^3_6*C^3_3\) by factorial of number of groups, so by \(2!\). Answer: B
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Re: M2416 [#permalink]
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03 Sep 2015, 00:57
Is there any easier way to do this, without using this formula? Sorry, was just wondering because I'm not used to seeing formulas like this, but if there is a trick or way to dumb it down, that would be awesome! If not, then can you tell me how I would step by step figure out the answer using this formula?



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Re: M2416 [#permalink]
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19 Sep 2015, 11:11
we have 6 boys. need to form two groups of 3 each.
so we need to see how many ways we can choose 3 from these 6 = 6!/3!*3! = 20 ways
so we can actually form 20 sets of 3 from these 6
and for the last part as bunel explained, in a group (a,b,c,d,e,f)  what we are doing in the above formula is finding ways in which groups of three can be made from these 6. so eg (a,b,c) will be one of those 20 and (d,e,f) will also be one of that 20.
now think about this in our scenario only 2 teams are formed. so if (a,b,c) are 1 team automatically the rest go to another team. so there is no need to count(d,e,f). same way if (d,e,f) is 1 team the rest are another team. so we can see that 20 has double counts. so we divide by 2. so ans is 10.



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Re: M2416 [#permalink]
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21 Dec 2016, 22:34
Hello, I'm very much confused about when to consider the order and when not to. I made two similar mistakes in this CAT which did affect the end result to 50. Please help me understand the order in permutation and combination questions.



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Re: M2416 [#permalink]
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22 Dec 2016, 01:53



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Re: M2416 [#permalink]
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22 Dec 2016, 10:16
Hi Buenei,
Can you please explain me the logic behind using 3C3?
Why have you multiplied with 3C3? What does 3C3 indicate?
And how do we solve such questions?



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Re: M2416 [#permalink]
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31 Dec 2016, 18:33
Hi Bunnel ,i have not understood the 3C3 part . Why do we need to multiply this 3C3 with 6C3/2! , Can you explain please .



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Re: M2416 [#permalink]
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06 Feb 2017, 17:13
sajib2126 wrote: Hi Bunnel ,i have not understood the 3C3 part . Why do we need to multiply this 3C3 with 6C3/2! , Can you explain please . To make it simpler, just imagine you were doing a game of 2v2. 4 dudes  Al, Ben, Chuck, Dan. How many ways can you choose two teams of two? (4c2)*(2c2)=6. You can only get {Al+Ben v Chuck+Dan}, {Al+Chuck v Ben+Dan}, and {Al+Dan v Ben+Chuck}. But wait! Because order does not matter, {Al+Ben v Chuck+Dan} is the same as {Chuck+Dan v Al+Ben} and you actually need to divide the 6 possibilities by the number of groups  2. Using math, (4c2)*(2c2)/2! = 3. In this problem, you have 6 dudes  Al, Ben, Chuck, Dan, Evan, Frank. Let's turn this into a game of 3 on 3. Okay, you need to choose 3 for one of the teams. 6c3=20. So, there are 20 ways to pick 3 dudes from a group of 6. Let's call this first group Group A. You have one team now and need to "create" another team. How many ways can you chose 3 dudes from the remaining 3 dudes? 3c3=1 way. Let's call this second group Group B.Now, think about the 20 ways you could pick 2 groups of 3 dudes from 6. If one team consists of Al, Ben, and Chuck in Group A, this is one of the 20 possible groups for Group A. This means that Group B (the other team) would have to be Dan, Evan, and Frank. [b] However, let's think about the other 19 possibilities. Picking Dan, Evan, and Frank in Group A is one of those other 19 possibilities, and one of the total of 20 possibilities. This would matter if the order mattered, but because you just want two teams  you don't really care which team is Group A and which is Group B , the order does not matter and these 20 choices actually have duplicates. So, divide 20 by 2 to get 10. In math speak, (6c3)*(3c3)/2!=10.



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Re: M2416 [#permalink]
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24 Feb 2017, 05:15
Hi,
you could also use the Basic counting principle.
For the first spot in group 1 you have 6 possibilities (6 boys) for the seconde 5 possibilities for the third 4 poss.. for the first in the second group 3 possi. for the second in the second group 2
So in total 6x5x4x3x2x1
Because the Order in the group doesn't matter (e.g. for Group 1 {A,B,C} = {B,C,A}) you have to divide bei 3! The Order of the Groups also doesn't matter ( G1, G2 = G2, G1) so you have to divide bei 2! again
in Conclusion:
\(\frac{6x5x4x3x2x1}{3!x2!}\)
Cheers



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Re: M2416 [#permalink]
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15 Aug 2017, 07:03
Hi Bunuel,
In case of 6C3*3C3, we are only doing the selection. If we had to do the arrangement we had to multiply this expression with 2!. I dont understand why we have further made a division by 2.



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Re: M2416 [#permalink]
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15 Aug 2017, 09:17
siddharthasthana2212 wrote: Hi Bunuel,
In case of 6C3*3C3, we are only doing the selection. If we had to do the arrangement we had to multiply this expression with 2!. I dont understand why we have further made a division by 2. We are dividing by \(2!\) as there are 2 groups and the order of the groups does not matter. For example consider the following 6 units: \(A\), \(B\), \(C\), \(D\), \(E\) and \(F\). Now, one of the groups that \(C^3_6\) gives is \(\{ABC\}\) and in this case the second group would be \(\{DEF\}\), so we have two groups \(\{ABC\}\) and \(\{DEF\}\). But \(C^3_6\) also gives \(\{DEF\}\) group and in this case the second group would be \(\{ABC\}\), so we have the same two groups: \(\{ABC\}\) and \(\{DEF\}\). Therefore to get rid of such duplications we should divide \(C^3_6*C^3_3\) by factorial of number of groups, so by \(2!\). Check other solutions of this question here: https://gmatclub.com/forum/howmanyway ... 05381.htmlSimilar questions to practice: http://gmatclub.com/forum/ninedogsare ... 88685.htmlhttp://gmatclub.com/forum/inhowmanyd ... 99053.htmlhttp://gmatclub.com/forum/6peopleform ... 95344.htmlhttp://gmatclub.com/forum/agroupof8 ... 55369.htmlhttp://gmatclub.com/forum/agroupof8 ... 06277.htmlhttp://gmatclub.com/forum/inhowmanyd ... 85707.htmlhttp://gmatclub.com/forum/howmanyways ... 05381.htmlhttp://gmatclub.com/forum/anthonyandm ... 02027.htmlhttp://gmatclub.com/forum/inhowmanyd ... 01722.htmlHope it helps.
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Re: M2416 [#permalink]
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06 Sep 2017, 23:55
Bunuel wrote: Official Solution:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)
A. 8 B. 10 C. 16 D. 20 E. 24
\(\frac{C^3_6*C^3_3}{2!}=10\), we are dividing by \(2!\) as there are 2 groups and the order of the groups does not matter. For example consider the following 6 units: \(A\), \(B\), \(C\), \(D\), \(E\) and \(F\). Now, one of the groups that \(C^3_6\) gives is \(\{ABC\}\) and in this case the second group would be \(\{DEF\}\), so we have two groups \(\{ABC\}\) and \(\{DEF\}\). But \(C^3_6\) also gives \(\{DEF\}\) group and in this case the second group would be \(\{ABC\}\), so we have the same two groups: \(\{ABC\}\) and \(\{DEF\}\). Therefore to get rid of such duplications we should divide \(C^3_6*C^3_3\) by factorial of number of groups, so by \(2!\).
Answer: B My answer was 20 because First group: 6B choose 3B, therefore 6!/(3!3!)=20 Second group: 3B left choose 3B, therefore there is only 1 was Since it´s an AND, 20*1=20 Please let me know where I am making a mistake, thank you!



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Re: M2416 [#permalink]
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06 Sep 2017, 23:59
giuliab3 wrote: Bunuel wrote: Official Solution:
How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)
A. 8 B. 10 C. 16 D. 20 E. 24
\(\frac{C^3_6*C^3_3}{2!}=10\), we are dividing by \(2!\) as there are 2 groups and the order of the groups does not matter. For example consider the following 6 units: \(A\), \(B\), \(C\), \(D\), \(E\) and \(F\). Now, one of the groups that \(C^3_6\) gives is \(\{ABC\}\) and in this case the second group would be \(\{DEF\}\), so we have two groups \(\{ABC\}\) and \(\{DEF\}\). But \(C^3_6\) also gives \(\{DEF\}\) group and in this case the second group would be \(\{ABC\}\), so we have the same two groups: \(\{ABC\}\) and \(\{DEF\}\). Therefore to get rid of such duplications we should divide \(C^3_6*C^3_3\) by factorial of number of groups, so by \(2!\).
Answer: B My answer was 20 because First group: 6B choose 3B, therefore 6!/(3!3!)=20 Second group: 3B left choose 3B, therefore there is only 1 was Since it´s an AND, 20*1=20 Please let me know where I am making a mistake, thank you! I tried to explain this in the discussion above, with examples and links to similar question. Please check. Hope it helps.
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Re: M2416 [#permalink]
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10 Sep 2017, 07:39
I took a different approach.
For any one person, we simply have to figure out how many possible different pairs of teammates they might have, which is 5*4/2 (/2 since order doesn't matter within the team).



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Re: M2416 [#permalink]
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28 Apr 2018, 21:51
+1 for option B. The catch here is the fact that grouping does not matter here.
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Re: M2416 [#permalink]
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29 Apr 2018, 17:59
I am also one of those that got caught by not dividing by 2!, but that is good to know. Good question










