six people are to be divided to three groups
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I'm going to assume that the question is asking how many ways 6 people can be divided into pairs.mariah wrote:If six people are to be divided to three groups, each of which contains two people, how many was are possible?
An easy approach would be to determine the number of choices we have for each person chosen:
6 people total.
The 1st person chosen can be paired with 5 different people, giving us 5 choices for the 1st pair.
6-2 = 4 people left.
The next person chosen can be paired with 3 different people, giving us 3 choices for the 2nd pair.
6-4 = 2 people left.
The next person chosen can be paired with only 1 person, giving us 1 choice for the 3rd pair.
Combining the number of choices we have for each pair:
(Number of choices for the 1st pair) * (Number of choices for the 2nd pair) * (Number of choices for the 3rd pair) = 5*3*1 = 15 possible pairs.
The approaches described in the posts above mine would be correct for the following problem:
Six people are to be divided into pairs and assigned to three different projects. Each project will be assigned a different pair, and no person is allowed to work on more than one project. How many different project assignments are possible?
Number of choices for 1st project = 6C2 = 15.
Number of choices for 2nd project = 4C2 = 6.
Number of choices for 3rd project = 2C2 = 1.
Combining, 15*6*1 = 90 possible project assignments.
Here's the difference: While 6 people can be divided into 15 pairs (the first problem), these 15 pairs can be assigned to 3 projects 90 different ways (the second problem).
Hope this helps!
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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