4 part combinitorial problem I believe this to be a 4 part combinatorics problem I am a programmer not a mathematician. The final answer I am looking for will be the total number of combinations for all 4 steps listed below
Step 1) 
There are 7 groups of objects, each group has 7 different objects in them (49 total unique objects).
You must choose 4 objects in each of the 7 groups for a total of 28 selections  (28 objects)
How many possible combinations of objects are there for all 7 groups in this step 1?
Based on this link:
https://math.stackexchange.com/a/383762/410036
This appears to be the formula for one group, the anser is 35 then to the 7th power because of 7 groups = 64,339,296,875 (i think?)
$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

Now I need to combine that with 3 more steps so the final number of combinations should be rather large.
Step 2) 
Now from the 4 items in each group that you chose in step 1 you must narrow down to only 1 in each group. (7 total objects selected)
Step 3)
Now from the 7 remaining objects selected (in Step 2) you must choose your 4 favorite.  (4 total objects selected)
Step 4)
Final step, just to make it fun, you must order the final 4 objects in order of your favorite 1,2,3,4.
I think this will be a very large number, any useful information regarding this problem is greatly appreciated.
Thank you.
Don
 A: Step 1 is correct. As stated by Ross Millikan, for step 2 you get $35^7*4^7$. For step 3, you must choose 4 out the 7 remaining items, so you get $C(7, 4) = 35$. Resulting in $35^7*4^7*35$. For step 4, start off with step 2 and do the same as in step 3, but use permutation instead of combination (as the order of the items is important). So $P(7, 4) = 840$, resulting in $35^7*4^7*35*840 = 30,991,570,176,000,000,000$. So indeed a huge number.
A: the below answer assumes that we are counting the number of possibilities and not the number of paths in which the process of selection happens. 
Step 1
Select $4$ objects from each of the $7$ groups So,
required ways $\dbinom{7}{4}^7$
Step 2
Similar approach
required ways $=\dbinom{7}{1}^7$
Step 3
Select 4 groups first, then select one person from each group.
required ways $=\dbinom{7}{4}\dbinom{4}{1}^4$
Step 4
due to ordering,
required ways $=\dbinom{7}{4}\dbinom{4}{1}^4(4!)$
A: You are correct for step 1.  Maybe you should report the result as $35^7$ instead of multiplying it out.  That is a matter of taste. Then is step 2 you choose ${4 \choose 1}$ seven times which gives another factor $4^7$ and so on.  Yes, it will be a very large number.
