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Re: How many ways are there to split a group of 6 boys into two [#permalink]

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06 Sep 2011, 12:04

if we select 3 boys in one group the remaining 3 boys will automatically fall into the other group.. also since the order does not matter... its just a matter of choosing 3 out of 6

therefore one can choose 3 boys out of 6 in 6c3 or 10 ways.

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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06 Sep 2011, 12:25

viks4gmat wrote:

if we select 3 boys in one group the remaining 3 boys will automatically fall into the other group.. also since the order does not matter... its just a matter of choosing 3 out of 6

therefore one can choose 3 boys out of 6 in 6c3 or 10 ways.

hence B

6c3 = 6!/3!3! i'm i on right ? and hence the answer is 20
_________________

How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground? - Jubran Khaleel Jubran

Last edited by Silver89 on 06 Sep 2011, 12:26, edited 1 time in total.

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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06 Sep 2011, 12:30

Silver89 wrote:

viks4gmat wrote:

if we select 3 boys in one group the remaining 3 boys will automatically fall into the other group.. also since the order does not matter... its just a matter of choosing 3 out of 6

therefore one can choose 3 boys out of 6 in 6c3 or 10 ways.

hence B

6c3 = 6!/3!3! i'm i on right ? and hence the answer is 20

I got 20 as well, but apparently it is not the correct answer. Expert Help is required !!

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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06 Sep 2011, 12:43

reatsaint wrote:

Silver89 wrote:

viks4gmat wrote:

if we select 3 boys in one group the remaining 3 boys will automatically fall into the other group.. also since the order does not matter... its just a matter of choosing 3 out of 6

therefore one can choose 3 boys out of 6 in 6c3 or 10 ways.

hence B

6c3 = 6!/3!3! i'm i on right ? and hence the answer is 20

I got 20 as well, but apparently it is not the correct answer. Expert Help is required !!

i guess because we have two groups and their order is not important so we need to divide by 2! 20/2=10 i just guessing
_________________

How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground? - Jubran Khaleel Jubran

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

Out of 6 boys, you can choose 3 in 6C3 ways and make the first group. The second group is of the remaining 3 boys. But when you do that, you have ordered the groups into first and second. The question mentions that the order of the groups does not matter. Hence, you need to divide your answer i.e. 6C3 by 2! to undo the ordering of the groups. That is, 6C3 gives you 20 ways. This includes G1 - ABC and G2 - DEF G1 - DEF and G2 - ABC as two different ways but it is actually just one way because there is no G1 and G2. In both the cases, the two groups are ABC and DEF Therefore, answer will be 20/2! = 10
_________________

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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07 Sep 2011, 11:18

reatsaint wrote:

viks4gmat wrote:

if we select 3 boys in one group the remaining 3 boys will automatically fall into the other group.. also since the order does not matter... its just a matter of choosing 3 out of 6

therefore one can choose 3 boys out of 6 in 6c3 or 10 ways.

hence B

6C3 is 20, not 10.

Thanks buddy... kicking myself now!! quite frankly now i dont know how i made such a bad mistake... got the answer 10.. checked the OA, was satisfied and posted my explanation... gosh hope i dont make such blunders on the real test

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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07 Sep 2011, 14:03

"This includes G1 - ABC and G2 - DEF G1 - DEF and G2 - ABC as two different ways but it is actually just one way because there is no G1 and G2. In both the cases, the two groups are ABC and DEF Therefore, answer will be 20/2! = 10"

I don't understand the explanation - If there are two groups ABC and DEF then how can there not be two groups?!

"This includes G1 - ABC and G2 - DEF G1 - DEF and G2 - ABC as two different ways but it is actually just one way because there is no G1 and G2. In both the cases, the two groups are ABC and DEF Therefore, answer will be 20/2! = 10"

I don't understand the explanation - If there are two groups ABC and DEF then how can there not be two groups?!

Say there are 6 boys: A, B, C, D, E, F There are two ways of splitting

Method I I start splitting them in two groups of 3 boys each. The two groups can be made in the following ways 1. ABC and DEF, 2. ABD and CEF, 3. ABE and CDF etc The groups are not names/distinct. If you have 6 boys in front of you and you split them in 2 groups and do not name the groups. 10 total ways.

Method II On the other hand, I could put them in two distinct groups in the following ways 1. Group1: ABC, Group2: DEF 2. Group1: DEF, Group2: ABC (If you notice, this is the same as above, just that now ABC is group 2) 3. Group1: ABD, Group2: CEF 4. Group1: CEF, Group2: ABD etc Here I have to put them in two different groups, group 1 and group 2. ABC and DEF is not just one way of splitting them. ABC could be assigned to group 1 or group 2 so there are 2 further cases. In this case, every 'way' we get above will have two possibilities so total number of ways will be twice. So there will be 20 total ways.
_________________

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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02 Nov 2011, 06:46

viks4gmat wrote:

if we select 3 boys in one group the remaining 3 boys will automatically fall into the other group.. also since the order does not matter... its just a matter of choosing 3 out of 6

therefore one can choose 3 boys out of 6 in 6c3 or 10 ways.

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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13 Nov 2011, 20:07

6C3 * 1 would give the different combinations of 3 boys from a group of 6. 6C3 = 20

If grouping mattered, then the answer would be 20. Here, 123 in G1 and 456 in G2 is different from 456 in G1 and 123 in G2.

But in our case, the grouping does not matter So 123 in a group (G) and 456 in a group(G) is same as 456 in a group(G) and 123 in a group(G). Hence, the right answer is 20 / 2 = 10.

I'd be interested if there is a formula way to calculate this kind of question. For example, if we were to make 3 groups of 2 people each (where groups do not matter), what would be the no. of ways to do so?

6C3 * 1 would give the different combinations of 3 boys from a group of 6. 6C3 = 20

If grouping mattered, then the answer would be 20. Here, 123 in G1 and 456 in G2 is different from 456 in G1 and 123 in G2.

But in our case, the grouping does not matter So 123 in a group (G) and 456 in a group(G) is same as 456 in a group(G) and 123 in a group(G). Hence, the right answer is 20 / 2 = 10.

I'd be interested if there is a formula way to calculate this kind of question. For example, if we were to make 3 groups of 2 people each (where groups do not matter), what would be the no. of ways to do so?

You already have the formula - nCr - It is pretty straight forward to solve these questions using it. To make 3 groups of 2 people each, you just do the following: You select 2 people out of 6 to make group 1, then 2 out of 4 to make group 2 and the last 2 are in group 3. 6C2 * 4C2 * 2C2/3! (you divide by 3! because the groups are not distinct)

Or another way to approach this is the following: Arrange all the people in a line which gives you 6! different arrangements. Now draw two perpendicular lines as shown to make 3 groups.

oo l oo l oo

Here, in each group, the people are ordered so to un-order them, divide by 2! three times. Also, the 3 groups are ordered so divide by 3! too. You get 6!/(2!*2!*2!*3!) which is the same as above.
_________________

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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14 Nov 2011, 23:53

VeritasPrepKarishma wrote:

You already have the formula - nCr - It is pretty straight forward to solve these questions using it. To make 3 groups of 2 people each, you just do the following: You select 2 people out of 6 to make group 1, then 2 out of 4 to make group 2 and the last 2 are in group 3. 6C2 * 4C2 * 2C2/3! (you divide by 3! because the groups are not distinct)

Or another way to approach this is the following: Arrange all the people in a line which gives you 6! different arrangements. Now draw two perpendicular lines as shown to make 3 groups.

oo l oo l oo

Here, in each group, the people are ordered so to un-order them, divide by 2! three times. Also, the 3 groups are ordered so divide by 3! too. You get 6!/(2!*2!*2!*3!) which is the same as above.

Thanks for the reply, Karishma. The second method was really awesome:-). I understand the logic now to convert from permutation to combination..or ordered to unordered....based on nCr = nPr / r!

Re: How many ways are there to split a group of 6 boys into two [#permalink]

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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

GENERAL RULE: 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 important is \(\frac{(mn)!}{(n!)^m}\)

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 NOT important is \(\frac{(mn)!}{(n!)^m*m!}\).

BACK TO THE ORIGINAL QUESTION: In original question as the order is NOT important, we should use second formula, \(mn=6\), \(m=2\) groups \(n=3\) objects (people): \(\frac{(mn)!}{(n!)^m*m!}=\frac{6!}{(3!)^2*2!}=10\).

This can be done in another way as well: \(\frac{C^3_6*C^3_3}{2!}=10\), we are dividing by \(2!\) as there are 2 groups and order doesn't matter.

For example if we choose with \(C^3_6\) the group {ABC} then the group {DEF} is left and we have two groups {ABC} and {DEF} but then we could choose also {DEF}, so in this case second group would be {ABC}, so we would have the same two groups: {ABC} and {DEF}. So to get rid of such duplications we should divide \(C^3_6*C^3_3\) by factorial of number of groups - 2!.

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