so I can think of 2 method but they give different answers which confuse me:
1st one is 48c6 x 42c6 x 36c6 x 30c6 x 24c6 x 18c6 x 12c6 x 6c6
each time you choose 6 people from the rest for 8 times
2nd one is 48!/((6!)^8 x 8!)
so choose 1 people from the rest for 48times, then divided by 6! 8 times since there are 8 groups and each groups has 6 people, to prevent duplicate counting since order doesnt matter, then divided by 8! since group order doesnt matter as well.
but this 2 methods have different answers, but both make sense to me, so which one is correct?
Best Answer
The problem with your first approach is that it badly overcounts. Suppose that your groups are A, B, C, D, E, F, G, and H. Your first calculation counts them $8!$ times: once when you choose them in the order ABCDEFGH, once when you choose them in the order ABCDEFHG, and so on, with a count for every possible order in which you could choose $8$ things. Since the overcounting is uniform, meaning that you’ve counted every set of groups the same number of times, you can easily correct for it: just divide your first result by $8!$.
You can also reason as follows. Imagine that we’ve numbered the people $1$ through $48$. There are $\binom{47}5$ ways to choose the other $5$ members of the group that contains person $1$. Let $k$ be the smallest number of a person not in that first group; there are $\binom{41}5$ ways to choose the other $5$ members of the group containing person $k$. Repeating this process six more times, we find that there are
$$\binom{47}5\binom{41}5\binom{35}5\binom{29}5\binom{23}5\binom{17}5\binom{11}5\binom55\tag{1}$$
ways to choose the $8$ groups. Note that
$$\binom{6n}6=\frac{6n}6\binom{6n-1}5=n\binom{6n-1}5\,,$$
so your product is indeed $8!$ times the product in $(1)$.