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Target Test Prep · GMAT

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Chris Peckover
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Oct 13 to Jan 7, 2027

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Logan Thompson
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Sep 6 to Dec 6, 2026

with Logan Thompson

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Sun · 9:30 AM to 12:30 PM ET
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Target Test Prep GMAT OnDemand

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GMAT PREP question-Permutation

Expert replies
Source: — Problem Solving |

by pepeprepa » Thu Aug 14, 2008 10:52 am
The total number of possibilities for the group of 3 is 8C3=56
Let's subtract the number of group in which there are couples.

In each group with a couple, the third one has 6 possibilities. So 6 possibilities of different groups for each couples. Given that there are four couples in total: 6*4=24

56-24=32
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by priyankamishra11 » Thu Aug 14, 2008 12:48 pm
Thanks i got it. :D
Regards,
Priyanka
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