# Two different results for splitting sample points into groups of a certain size?

Let's say I've got a group of 12 people and want to split them into groups of 4.

According to this post How many ways $12$ persons may be divided into three groups of $4$ persons each?, the possible number of combinations is

$$\frac {12!}{(4!4!4!3!)} = 5775$$

However, to my knowledge, the number of possible combinations of n things taken r at a time is

$$\frac{ n!}{(r!(n-r)!)}$$,

so in this case $\frac{12!}{(4!8!)} = 495$ .

What is the difference and why does this occur?

I know that I am somehow making a mistake and have misunderstood something completely, but I can't make sense of it!

$n!/(r!(n-r)!)$ counts the number of ways to form a single group of $r$ people out of a pool of $n$ people. This is a different problem from dividing the people into three groups.

There are $\binom{12}{4}=\frac{12!}{4!8!}$ ways to select $4$ persons out of twelve, but this is not the same thing as splitting twelve people into groups of four, we can view this as selecting a group of four people from twelve.

One way to separate twelve people into groups of four is to do this three times. To do this select team number $1$ in $\binom{12}{4}$ ways. Then select team number two from the remaining $8$ people in $\binom{8}{4}$ ways and then select team number three from the remaining 4 in $\binom{4}{4}=1$ ways. However we don't care which number each team has. We have counted each result $6$ times depending on which number each team gets. Therefore the final result is:

$$\dfrac{\binom{12}{4}\binom{8}{4}\binom{4}{4}}{3!}=\frac{12!}{8!*4!}\frac{8!}{4!*4!}\frac{4!}{4!*1}*\frac{1}{3!}=\frac{12!}{4!*4!*4!*3!}$$

The original question was asking about How many ways 12 persons may be divided *into three groups of 4 persons each*?. That's not the same as how many possible groups of 4 people can be formed from 12.

Another way to look the actual answer $\frac{12!}{4!4!4!3!}$ is to see the $12!$ as simply the initial ordering for the whole group. Then cutting the three groups arbitrarily (first four, middle four, last four), we see that each of those groups has been counted $4!$ times, and given that the groups are interchangeable, the first group has also been counted when it was the middle group, etc. for another factor of $3!$ to divide out.

Note that if we had been dividing the people into the distinguishable teams - red, blue and yellow, say - the final dividing factor of $3!$ would not have applied.