How many ways may 9 different chocolate bars be given to 3 children’s such that each child gets 3 chocolate bars? Assume order is important How many ways may 9 different chocolate bars be given to 3 children’s such that each child gets 3 chocolate bars? Assume order is important. 

I’m not sure how to solve this question or where to even start ive tried a few ways but don’t get the correct answer
There are $\binom{ 9}{3, 3, 3} = \frac{ 9!} { 3! 3! 3!} = 1680 $ ways to distribute 9 distinct chocolate bars to the 3 children, each receiving 3 of them. This is the Multinomial theorem.

If you care about the order, then you can multiply by the order that you care about (IE order just to an individual child, or order of all 9).
It seems to me that the division among the three children is just dust in the eyes if the order is important. Let me unpack. 

Let's call the chocolate bars: 1, 2, ..., 9.

Suppose given a list of these bars you give the first three to the first kid, second three to the second kid, last three to the third kid. Let's depict it this way:

c1 c2 c3 | c4 c5 c6 | c7 c8 c9,

where ci is the ith chocolate bar. If the order is important, i.e.

1 2 3 | 4 5 6 | 7 8 9

and 

2 1 3 | 4 5 6 | 7 8 9

count as two different ways of giving out bars, then the question really is: "How many permutations of the 9 chocolate bars exist?". And the answer is 9!.

If this is not the answer, then the question should be better posed.