how many ways to arrange 10 players between 3 positions with at least two players in each position? my thought is that you would do 10 choose 6 to decide the 6 players to split between the 3 groups, then 6 choose 2 times 4 choose 2 times 2 choose 2 divided by 3! to determine how to arrange the 6 and then finally 3^4 for the remaining players but this number is larger than the total number of ways to arrange the players which is 3^10?

One straightforward way to do it is to start by listing the possible breakdowns of the numbers of people at three positions: $\langle 4,3,3\rangle$, $\langle 4,4,2\rangle$, $\langle 5,3,2\rangle$, $\langle 6,2,2\rangle$. Now handle these cases one at a time. I’ll do the first one.

$\langle 4,3,3\rangle$: There are $\binom{10}4$ ways to choose the $4$-person position, and then $\binom63$ ways to choose the first of the remaining positions, for a total of $\binom{10}4\binom63$ ways to choose a group of $4$ people and a first group of $3$ people. The remaining $3$ people will of course form the third group. But there are $3!=6$ ways to permute the three groups, so the final total for this case is $6\binom{10}4\binom63$.

The reason that your approach doesn’t work is that you’re counting some arrangements more than once. Suppose that the players are $P_1,P_2,\ldots,P_{10}$, the $6$ that you choose initially are $P_1$ through $P_6$, you split them $\{P_1,P_2\},\{P_3,P_4\}$, and $\{P_5,P_6\}$, and then you add the remaining four to the first group. You now have $P_1,P_2,P_7,P_8,P_9$, and $P_{10}$ at the first position, $P_3$ and $P_4$ at the second position, and $P_5$ and $P_6$ at the third position. But you also get that arrangement if your first $6$ are $P_3,P_4,P_5,P_6,P_7$, and $P_8$, you put $P_7$ and $P_8$ in the first position, $P_3$ and $P_4$ at the second, and $P_5$ and $P_6$ at the third, and then put all $4$ of the remaining players at the first position. In fact, there are $\binom62=15$ different ways to reach this same arrangement, so you’re doing a lot of overcounting.
