Group Combinations Problem

**mathceleb** · July 3rd 2010, 10:15 AM

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** · July 3rd 2010, 11:17 AM

Originally Posted by mathceleb

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 $\displaystyle \frac{1}{\binom{149}{2}}$ , 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 $\displaystyle \binom{149}{2}$ ways to choose two people out of the remaining 149...

**mathceleb** · July 3rd 2010, 12:38 PM

I'm not sure. I found this thread from back in the day which is a similar question...

http://www.mathhelpforum.com/math-he...mutations.html

**Soroban** · July 3rd 2010, 01:16 PM

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: . $\dfrac{150!}{(3!)^{50}}$ possible groupings.

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

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

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

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

**Plato** · July 3rd 2010, 01:37 PM

Originally Posted by undefined

the probability is $\displaystyle \frac{1}{\binom{149}{2}}$ , 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 $\displaystyle \binom{149}{2}$ ways to choose two people out of the remaining 149...

This correct.

Originally Posted by Soroban

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

Those are ordered partions.

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

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

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

**undefined** · July 3rd 2010, 01:45 PM

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

Originally Posted by Soroban

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

I believe this should be $\dfrac{\left(\dfrac{150!}{(3!)^{50}}\right)}{50!}$ because the groups of 3 can be permuted.

Originally Posted by Soroban

...Hence, there are: . $\dfrac{147!}{(3!)^{49}}$ ways to have the 3 blue people in one group.

Likewise I believe this should be $\dfrac{\left(\dfrac{147!}{(3!)^{49}}\right)}{49!}$ .

Leading to:

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

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.

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