Customizing Modular Storage Files: 560 Options!

[Hodgey8806]

Homework Statement

An office furniture manufacturer that makes modular storage files offers its customers two choices for the base and four choices for the top, and the modular storage files come in five different heights. The customer may choose any combination of the five different-sized modules so that the finished file has a base, a top, and one, two, three, four, five, or six storage modules.

How many choices does the customer have if the completed file has four storage modules, a top, and a base? The order in which the four modules are stacked is irrelevant.

Homework Equations

Really quick, I know that the answer is 560. But I don't know how I can get there. I'm doing a bit of backwards reasoning because I can't find the missing multiple:

The easy part is that there 2 parts I know easily: 2 choices for the bottom, and 4 for the top. So that is already a multiplier of 8 for the missing multiple. So the equation is as follows:

2*4*( ) = 560

So, the missing piece needs to be 70.

The Attempt at a Solution

To find this 70, my reasoning was that we have 5 size choices for the middle 4 module. But order is irrelevant so instead of a multiplier rule (5^4), we need a multinomial probability.

The only way I know to do that is consider:
What if all were different? Isn't that 120 [ ie (5!)/(1!1!1!1!1!) ]choices?

What if none were different? [(5!)/(4!1!)] = 5.
What if 1 is different? [(5!)/(3!1!1!)] = 20
What if two were are the same: i) But two are different: [(5!)/(2!1!1!1!)]=60
ii) But two are the same(two pair): [(5!)/(2!2!1!)]=30
What if 3 were different? Then it'd be the same and it follows that they are all different, correct?
--and if 3 were the same, isn't that the same as 1 is different. So it is used up by the 1 being different.

And if 4 were different, isn't that the same as 3 different?

If you can help me find the "considerations" I need to make, it would be great! I need to know how to get 70 because none of those do unless I just pick 1 different or 2 different pair (20+30 = 50) and from all different: 120-50 = 70.

Now if that is the correct way to do it, I need to see why this is acceptable. Thanks!

 

[vela]

Say you used two different sized modules A and B. With those two, you can make three distinct combinations: AAAB, AABB, and ABBB, so many combinations can you get in total of two differently sized modules?

 

[Hodgey8806]

6? 3 combinations per module. Am I thinking correctly?

 

[vela]

I think you didn't understand what I was asking. The question wasn't very clear. You have five different sizes. For each pair of sizes you can choose, you can make the three distinct combinations of four modules, so how many combinations in total do you get?

 

[Hodgey8806]

I got 30. Because 3 distinct combinations for AB, AC, AD, and AE = 12 for using A
Then BC, BD, BE = 9
Then CD, CE = 6
Then DE = 3

I think that's right. But, isn't that what the (5!)/(2!2!1!)

 

[vela]

That's right, but the 30 total doesn't correspond to what you described as the number of combinations consisting of two pairs. ABBB and AAAB don't consist of two pairs.

The number of possible pairs is 5C2=10, right? And then for each pair, you can make three distinct configurations, so you get a total of 3x10=30.

Now what do you get for combinations like ABCC, and ABCD? You already counted the AAAA-type combos correctly in your original post.

 

[Hodgey8806]

Combination like ABCC I get 30 again I believe. If I did it right. Is it the same factorial but with different meaning? I am seeing it now as (5!)/(2!2!1!) In which 2 are the same, but 2 are different.

And then for ABCD, I must have misunderstood the multinomial factorial. Is it (5!)/(4!1!) Where 4 is the number of different ones? If so then it does =70!

 

[Hodgey8806]

That's an excited 70, not a 70 factorial lol

 

[vela]

Combination like ABCC I get 30 again I believe. If I did it right. Is it the same factorial but with different meaning? I am seeing it now as (5!)/(2!2!1!) In which 2 are the same, but 2 are different.

I'm not sure how to apply the multinomial coefficient to this problem, and I'm too lazy to figure out if you can. Can you explain your interpretation of 5!/(2! 2! 1!)? When you said "2 are the same, but 2 are different," it didn't really mean anything to me.

The way I'm looking at it is: You need three different sizes, so there are 5C3 ways to select those. Then one of the sizes is repeated, so there's three ways to do that, i.e. AABC, ABBC, or ABCC.

And then for ABCD, I must have misunderstood the multinomial factorial. Is it (5!)/(4!1!) Where 4 is the number of different ones? If so then it does =70!

For ABCD, you need 4 chosen out of 5, which gives you that binomial coefficient.

 

[Hodgey8806]

The two are the same means that there are two pair, and then there are two that aren't a pair but we don't care about their order such as AABC = AACB so we divide by 2! to take the order out.

 

[Hodgey8806]

I found the formula I was meant to use. The book talks about it without how to use it--they said it's unnecessary for probability. It is for a sample space r for n when we don't care about order and we sample with replacement.

It is the formula: (n-1+r)C(r) or (n-1+r)C(n-1)

If we apply it, it will give me 70 which results in the 560 that I need for the answer.