(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant 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.

3. 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!

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# Homework Help: Discuss finding the number of choices a customer has using multinomial probability.

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