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Re: M06-11 [#permalink]
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Another way to think about this is remove the:
- meaning for order within the groups
- meaning for order among the groups
solution:
1. we arrange all the 8 people in a line, this gives us: 8!
2. since the order among the gourps does not matter, we will have to devide 8! by 4!
3. since the order within the groups allso does not matter, for each group, we will have to devide by 2!

hence we got: 8!/(4!2!2!2!2!) = 105

comment: by saying "the order does not matter" i mean that, for example if we take the order within the gourp, it does matter if i picked person X to be the first in the group and later person Y to be in the group.
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Re: M06-11 [#permalink]
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I had trouble conceptualizing this problem at first, and after re-visiting Combinatorics theory and Bunuel's response I realized the best way I understood it was as having 2 sets of combination selections: 1) Selecting players per each team of 2, 2. Selecting 4 teams. Hence that is why the Combination formula needs to be applied twice.

Illustrating this:

1) Selecting players per each team of 2:

Out of 8 players, selecting 1st team:
\(\frac{8!}{(8-2!)2!}\) = 28 or else \(\frac{8*7}{2!}\) = 28

Since we have selected 2 out of 8 players, that leaves 6 possible players for 2nd team:
\(\frac{6*5}{2!}\) = 15

3rd team:
\(\frac{4*3}{2!}\) = 6

4th team:
\(\frac{2*1}{2!}\) = 1

2) Selecting 4 teams (Combination instead of Permutation since order does not matter):

\(\frac{(1st team * 2nd team* 3rd team * 4th team)}{4!}\)

= \(\frac{(28*15*6*1)}{4!}\)

= 105
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Re: M06-11 [#permalink]
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i solved this question in visualization manner

In first pick , I choose
BF-->DG -->AC -->EH

Means in first pick i had 28 options ; in 2nd pick i had 15 options, in 3rd pick i had 6 options and hence in the last only 1 option

it is same way as saying:
Pick 2 out of 8; then pick 2 out of remaining 6 ; pick 2 out of remaining 4 and then 2 out of 2/


But i need to divide by 4! because I had 4 teams without any order.
it means :
I got the combinations included:
BF - DG - AC - EH
BF -DG - EH - AC
etc
24 such combinations are part of 28*15*6*1
But as order doesnt matter so I will divide it by 24

hence final answer = 28*15*6/24 = 105
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Re: M06-11 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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M06-11 [#permalink]
8!/(2!*2!*2!*2!*4!)

It gives 105.

P.d.
if you cross out the numbers while simplifying, make sure that you are crossing out very well, if not you may end up multiplying by a number that had been eliminated already. For example, if you crossed out 4 and then thought that it is still there, you may have multiplied by 4 and got 420.­ And guess what 420 is an answer choice.
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M06-11 [#permalink]
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