How many ways are there to place nine people in three boats, if in each boat we should place 3 people? The task is:

How many ways are there to place nine people in three boats, if in each boat we should place 3 people.

1) $\displaystyle \frac{9!}{3!\times6!}=84$ different triplets
2) $\displaystyle \frac{84!}{3!\times(84-3)!}=95284$
But the answer is $1680$.
What did I wrong?
 A: We pick $3$ people out of the $9$ for the first boat, out of the remaining $6$ people we choose $3$ people for the second boat and finally the 3 remaining people go in the last boat. Hence there are
$$
\binom{9}{3}\binom{6}{3}\binom{3}{3}=\frac{9!}{3!3!3!}=1680
$$
ways.
A: What you did doesn't work because these $84$ different triplets include all possible triplets with three different people. For example $(1,2,3)$, $(1,2,5)$ and $(1,3,5)$. 
Now with the second calculation, $\dfrac{84!}{3!(84-3)!}=95284$, you calculate the number of ways to choose three of these  $84$ different triplets. However, you might choose $(1,2,3)$, $(1,2,5)$ and $(1,3,5)$. Then you don't divide all people between the boats. You have to choose three triplets that contain people one through nine exactly once, and this calculation fails to take this into account. 
The correct calculation is in the other answer by Foobaz John. 
A: Assume the three boats are $b1, b2, b3$.  The number of ways to choose 3 people for boat $b1$ is ${9 \choose 3} = 84$. The number of ways to choose 3 people for boat $b2$ is ${6 \choose 3}=20$ because we can choose any members for this boat except the 3 people who have already been chosen for $b1$.  The members of the third boat are now uniquely determined to be the remaining 3 people.  So the number of ways to place 9 people in 3 distinguishable boats is ${9 \choose 3} \cdot {6 \choose 3} = 84 \cdot 20 = 1680$.
A: Another way to arrive at
$$\frac{9!}{3!3!3!}$$
is to imagine lining up the nine people in all possible orders (9!); the first three go in the first boat, next three in the second boat, etc. But within each boat, their order doesn’t matter, so we divide by 3! per boat. I find this more intuitive, myself, and easier to apply to more complex cases. For example, distribute 10 people among boats of sizes 3, 2, and 5: $10!/3!2!5!$
