Arranging 3 items in 7 spots I just wanted to make sure my answer to a problem is correct. 
If we have 3 different items, how many ways can we arrange them amongst 7 spots? What if each item was exactly the same and we didn't care about the order of the 3 items?
If we always pick the same item first and so on, we get 7 * 6 * 5 because for the first item we have 7 possible spots then 6 then 5. Then by symmetry we multiply by 3! to account for the number of possible permutations of the items to get 7!3!/4!? Then we could divide by 3! to drop off all permutations of items to get 7!/4! .
Another way I though about it was we have 7 total items, one group of 3 and one of four. 7!/4! would drop all permutations of the group of 4 which would be a sort of parallel to having 3 items in 7 spaces where we care about the order. From that we could drop all permutations of 3! to get 7!/4!3!.
The numbers seem off. Where did I go wrong? I'm guessing somewhere with the 7!3!/4!  
 A: This answers the question posed in your first comment under the question.
First for convenience let's make it smaller:
Suppose there are $2$ items $b,c$ and $3$ spots. First I pick $b$ and place it on one of the $3$ spots. Possible result:


*

*$b..$

*$.b.$

*$..b$


Now I take $c$ and place it on one of the remaining spots. Possible results:


*

*$bc.$ or $b.c$

*$cb.$ or $.bc$

*$c.b$ or $.cb$


If I permute $b$ and $c$ then will I get any results that were not allready there? No! So multiplying by $2!$ would be wrong here.
A: If you assume the 3 items are same, then it is not possible to order them.
So the answer is simply in how many ways you can select 3 spots out of the 7.
Which is ${7}\choose3$=$\frac{7!}{3!4!}$.
A: If we assume all three items are distinct then there are 7 places you can place the first item, 6 places remaining to place the second item and finally 5 places to place the final item making 
$$7 \times 6 \times 5 = 210 \text{ distinguishable solutions}$$
Now if the items are not distinct there will be fewer distinguishable solutions not more. We can organise 3 items in $3 \times 2 \times 1 = 3! = 6$ ways so we divide by 3! not multiply giving:
$$\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = \dfrac{210}{6} = 40 \text{ distinguishable solutions}$$ 

To try and explain further lets say there are three balls red, green and blue as you have already worked out this gives you 210 permutations.  Now imagine a blind man who can not see the balls but can feel them.  To him because the balls all feel the same and we can arrange 3 objects in $3! = 6$ ways you can distinguish 6 times the number of solutions than he can. Because you can see the colours while he can not.
Because of this we divide by 6 instead of multiply.
