Expert: Josh - 8/29/2005

Question
Hi again Josh!

[Question]
There are 5 people, 3 republicans and 2 democrats. There is a table
that will seat 3 people.

a) How many combinations are there of arranging the groups at the
table.

b) Although order does not matter, how many permutations are there.

[Difficulty]
I know the answers really, I just don't know how you would go about getting them/what formula to use, because somthing like 5C3 wouldn't work.

I understand BASIC combinations and permutations and the formulas...

[Thoughts]
a)

3

RRR, RRD, DDR

b)

7

RRR, RRD, RDR, RDD, DRR, DDR, DRD

Thanks for your time!:)

Answer
It's really good that you are showing your thoughts. I would approach the problem this way.
Let R=3 for the republicans group, D=2 for the democrats.
We have to choose j from R and k from D, such that j+k=n and k<=R, for j=0,1,...,min(n,R). Here, n=3.

Since k cannot exceed two (k<=2), one person must be a republican.
We first sit this republican at the table.
Next, we have to sit one person on either side of this republican.
One way to do this is to surround the republican with two R's.
* Another way is to sit a "D" on the left, a "R" on the right.
**We can also reverse this, and sit a "R" on the left, a "D" on the right.

All in all, there are 3 combinations.

"*" can be ordered in C(3,2), while "**" can be ordered in C(3,1) ways. Adding these up gives 7 permutations -- if order matters.

Cheers:)