# Basic combinations?

• Sep 15th 2009, 05:13 PM
elellilrah
Basic combinations?
Here's an interesting toy problem:
You have 20 people sitting around a table. You want to form a committee of 3 people.
What is the probability/chance that at least 2 of the people are sitting together? What is the general form of the approach to finding the answer for this one?
It's been 15 years since I've done stats and surely this is easier than I'm making it out to be.
Thanks!
Aaron
• Sep 15th 2009, 05:45 PM
Soroban
Hello, Aaron!

Quote:

You have 20 people sitting around a table.
You want to form a committee of 3 people.

What is the probability/chance that at least 2 of the people are sitting together?

Suppose the people are: .$\displaystyle \{A,B,C,\hdots, S,T\}$
. . seated around the table in that order.

There are: .$\displaystyle _{20}C_3 \:=\:1140$ possible committees.

We must consider two cases:
. . (1) The 3 people are sitting together.
. . (2) Exactly 2 of the 3 people are sitting together.

(1) 3 people are together.
There are 20 ways: .$\displaystyle \{ABC,\:BCD,\:CDE,\:\hdots\:RST,\:STA,\:TAB \}$

(2) Exactly 2 people are together.
There are are 20 choices for the Pair: .$\displaystyle \{AB,\:BC,\:CD,\:\hdots\:ST,\:TA\}$

There are 18 choices for the third person.
But he/she must not be adjacent to the Pair.
Hence, there are only 16 choices for the third person.

There are: .$\displaystyle 20 \times 16 \:=\:{\color{red}320}$ ways.

Therefore, there are: .$\displaystyle 20 + 320 \:=\:340$ ways.

. . $\displaystyle P(\text{at least 2 together}) \:=\:\frac{340}{1140} \:=\:\frac{17}{57}$

• Sep 16th 2009, 10:05 AM
elellilrah
response
Soroban - Thank you very much for the quick answer and clear discussion of this question. It'll help immensely comparing this direct answer with the numerical simulation we're running.

Best regards,
aaron