Group Combinations Problem

Feb 2008
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Berkeley, Illinois
If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

Any thoughts?
 

undefined

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Mar 2010
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If we have a group of 150 people, broken into groups of 3, and only 3 are blue, what is the probability that all 3 blue people are put in the same group?

Is the denominator 150 C 3 = 551,300? For the numerator, my guess was 3!?

Any thoughts?
I haven't worked this out rigorously, but I think the probability is \(\displaystyle \displaystyle \frac{1}{\binom{149}{2}}\), by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are \(\displaystyle \displaystyle \binom{149}{2}\) ways to choose two people out of the remaining 149...
 

Soroban

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May 2006
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Hello, mathceleb!

We have a group of 150 people, broken into groups of 3, and only 3 are blue.
What is the probability that all 3 blue people are put in the same group?

The 150 people are divided into 50 groups of 3 people each.

There are: .\(\displaystyle \dfrac{150!}{(3!)^{50}}\) possible groupings.


To have the 3 blue people in one group, there is: .\(\displaystyle {3\choose3} = 1\) way.

The other 147 people are divided into 49 groups of 3: .\(\displaystyle \dfrac{147!}{(3!)^{49}}\) ways.
. . Hence, there are: .\(\displaystyle \dfrac{147!}{(3!)^{49}}\) ways to have the 3 blue people in one group.


The probability that all 3 blue people are in one group is:

. . \(\displaystyle \dfrac{\dfrac{147!}{(3!)^{49}}} {\dfrac{150!}{(3!)^{50}}} \;=\;\dfrac{147!}{(3!)^{49}}\cdot\dfrac{(3!)^{50}}{150!} \;=\;\dfrac{6}{150\cdot149\cdot148} \;=\;\dfrac{1}{551,\!300} \)

 

Plato

MHF Helper
Aug 2006
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the probability is \(\displaystyle \displaystyle \frac{1}{\binom{149}{2}}\), by reason that: consider blue person #1 (label them arbitrary 1,2,3); blue person #1's two team members are equally likely to be the two other blue people as any other two people. There are \(\displaystyle \displaystyle \binom{149}{2}\) ways to choose two people out of the remaining 149...
This correct.


The 150 people are divided into 50 groups of 3 people each.
There are: .\(\displaystyle \dfrac{150!}{(3!)^{50}}\) possible groupings.

Those are ordered partions.

There are \(\displaystyle \dfrac{150!}{(3!)^{50}(50!)}\) unordered partitions.

There are \(\displaystyle \dfrac{147!}{(3!)^{49}(49!)}\) of those where the blues are together.

Note that \(\displaystyle \dfrac{\dfrac{147!}{(3!)^{49}(49!)}}{ \dfrac{150!}{(3!)^{50}(50!)}}=\dfrac{1}{\binom{149}{2}}\)
 

undefined

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Mar 2010
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Chicago
Hmm since Soroban's answer differs from mine I decided to do a little more research.


There are: .\(\displaystyle \dfrac{150!}{(3!)^{50}}\) possible groupings.
I believe this should be \(\displaystyle \dfrac{\left(\dfrac{150!}{(3!)^{50}}\right)}{50!}\) because the groups of 3 can be permuted.


...Hence, there are: .\(\displaystyle \dfrac{147!}{(3!)^{49}}\) ways to have the 3 blue people in one group.
Likewise I believe this should be \(\displaystyle \dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)}{49!}\).

Leading to:

\(\displaystyle \dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)\cdot50!}{49!\cdot\left(\dfrac{150!}{(3!)^{50}}\right)} = 50\left(\dfrac{1}{551,\!300}\right) = \dfrac{1}{11026} = \displaystyle \frac{1}{\binom{149}{2}}\)

See here (a PDF file) for some more info; under subheading "Partitioning".

Edit: Plato answered a lot quicker than I did, didn't refresh the page.