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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

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    Wow! I did not expect answers so fast. I am reading them now. – 2017-01-25
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    I belive the possiblities vs paths question is ansered by my inclusion of : "Now from the N items in each group that you chose in previous step?" It is the "number of paths" if my understaing is corect. I like math. – 2017-01-25

3 Answers 3

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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.

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    Wow, I think you are quite brilliant. It seems so simple the way you explain it. – 2017-01-25
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    Something that helps me to solve combinatorial exercise (apart from making a lot of exercises and discussing them), is to think in terms of two questions: "Is the order of items important or not?" and "Can you use one item several times (repeat items)?". Just a little method I think is a good approach to these exercises :-) – 2017-01-25
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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 $=\dbinom{7}{4}\dbinom{4}{1}^4(4!)$

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    I don't think you have to divide with 4!. That over counting is already taken into account with the combination definition: $C(7, 4) = \frac{7!}{(4!*(7-3)!)}$ – 2017-01-25
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    @mitchbus, you are right, no over-counting occurs and hence no need to divide. – 2017-01-25
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    @Kirian, I sure do appreciate you explaining each step. I very much thank you. – 2017-01-25
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    @DonMoore, welcome. – 2017-01-25
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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.

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    you mean, in step 2, $35^7 \times 4^7$ possibilities are there? – 2017-01-25
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    That is right. That is the number of ways to get to where you are at that point. It remembers the items you chose in step 1 and threw away in step 2. – 2017-01-25
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    I think the OP tries to count all the possible paths to select the items, rather than the possible items – 2017-01-25
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    selecting first 4 persons, and then selecting one from these 4 gives duplicates. i think, this is equivalent to selection one from from 7 and my answer was written in this belief. – 2017-01-25
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    @mitchbus, oh got the point. my entire answer was based on the assumption that we are counting the number of possibilities and not the number of paths. i have now added this in my answer. – 2017-01-25
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    @Kiran: it gives duplicates in the list of persons selected. It does not give duplicates in the route taken to select those people. I think the question clearly wants the latter. – 2017-01-25
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    @RossMillikan, yes, i got the point. thanks for explaining. – 2017-01-25