# Group Combinations Problem

#### 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?

#### undefined

MHF Hall of Honor
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

MHF Hall of Honor
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
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

MHF Hall of Honor
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.

Similar threads