# Data Help!!! Counting Techniques

• Mar 3rd 2008, 02:08 PM
wikji
Data Help!!! Counting Techniques
A class has 12 students. In how many different ways can the students be put into groups of 3?

Answere is 369 600
• Mar 3rd 2008, 02:53 PM
roy_zhang
Since the order of students within each group does not matter, we use combination as:

$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}= 369600$
• Mar 3rd 2008, 03:33 PM
Plato
Quote:

Originally Posted by wikji
A class has 12 students. In how many different ways can the students be put into groups of 3?
Answere is 369 600

That is not the answer to the problem as it is stated.
That is the answer to this question: “A class has 12 students. In how many different ways can the students be put into teams of 3; a red team, a green team, a blue team, a brown team?” Someone might prefer being on the red team rather than being on a blue team. These are known as ordered partitions.

But that is not the question as stated.
This is like dividing the class into ‘study-groups’ where only content matters. These are known as unordered partitions.
The answer in this case is $\frac{{\left( {12} \right)!}}{{\left( {3!} \right)^4 \left( {4!} \right)}}=15400$