We need to make 2 different groups containing 2 each from 4 people {x,y,c,z}
- (x,y) and (c,z)
- (x,z) and (y,c)
- (x,c) and (y,z)
so we have 3 options
but when we use the formula C(4 2), we get 6. doesn't the combination formula eliminates the possibilities when switch places for who we choose first in each group?
(y,x) and (z,c)
(z,x) and (c,y)
(x,c) and (z,y)
The combination formula $C(4,2)$ tells you how many groups of 2 we can form from 4 objects. So, $(x,y)$ is one group, $(x,z)$ is another group, and so on. The six possibilities are $(x,y), (x,c), (x,z), (y,c), (y,z), (c,z)$.
Your question is slightly different. What you want to do is take 4 people and put them into two groups of two people, where the order of the groups doesn't matter. One way to solve this is as follows (but there's probably a simpler way...): The multinomial formula $\displaystyle \frac{n!}{n_1! n_2! \cdots n_k!}$ tells us how many ways there are to partition $n$ objects into $k$ bins of sizes $n_1, n_2, ..., n_k$ where $n_1 + n_2 + ... + n_k = n$. For our case, we have $n = 4$, $n_1 = 2, n_2 = 2$. But this formula counts each bin as distinct, so the first bin of size 2 is different from the second bin of size 2. For example, if we have $(x,z)$ in bin 1 and $(x,y)$ in bin 2, this is different than having $(x,y)$ in bin 1 and $(x,z)$ in bin 2. In your case, however, you don't make such a distinction, so to remove the repeats we divide by the number of ways to arrange the two bins in a row, which is just 2! = 2. The final answer gives $\displaystyle (\frac{4!}{2!2!}) / (2!) = 3$.