# Combinations. Confused on the terms used in a solution.

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• Jun 14th 2012, 08:26 PM
polcareb192
Combinations. Confused on the terms used in a solution.
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.

The Answer in the appendix is 560.

I found a solution which had 5-1+4C4 (4) (2) = 560. I do not understand the terms. Why is it 5-1+4C4?
• Jun 14th 2012, 08:27 PM
polcareb192
Re: Combinations. Confused on the terms used in a solution.
This example is from Hogg and Tanis Probability and Statistical Inference, eighth edition.
• Jun 14th 2012, 08:39 PM
ANDS!
Re: Combinations. Confused on the terms used in a solution.
Something is off in your original post: 4C4 is 1. How are you getting 560 out of that? And do they have 6 choices for storage or 5?
• Jun 14th 2012, 11:11 PM
richard1234
Re: Combinations. Confused on the terms used in a solution.
You have four storage modules, and each one can be one of five heights. The thing is, ordering is irrelevant. Assume that the heights are 1,2,3,4,5 (1 is shortest). We have to use some casework:

Case 1: AAAA (all four modules have the same height).
This can occur five ${{5}\choose{1}} = 5$ different ways (1111, 2222, ..., 5555).

Case 2: AAAB (three with the same height, one differs)
This can occur $2 {{5}\choose{2}} = 20$ ways. We must choose two numbers from 1,2,3,4,5, and choose which one to use three times.

Case 3: AABB (two pairs of two modules of equal height)
This occurs ${{5}\choose{2}} = 10$ ways.

Case 4: AABC (two with the same height, two with different heights)
This occurs $3 {{5}\choose{3}} = 30$ ways. Choose three numbers from {1,2,3,4,5}, choose the one to repeat.

Case 5: ABCD (all four have different heights).
This occurs ${{5}\choose{4}} = 5$ ways.

Therefore the total number of height arrangements is 5+20+10+30+5 = 70. There are two choices for the base and four for the top, so the total number of possible completed storage files is 70*2*4 = 560.
• Jun 15th 2012, 08:41 AM
polcareb192
Re: Combinations. Confused on the terms used in a solution.
Thank you Richard. The solution I found was supposed to read (5-1+4)C4 (2)(4) --- It ends up as 8C4 X 2 X 4 = 70 X 2 X 4 = 560. Your solution is more thoroughly explained and simpler. Thanks!
• Jun 15th 2012, 01:01 PM
richard1234
Re: Combinations. Confused on the terms used in a solution.
No problem :)
• Jun 18th 2012, 11:14 PM
psrose
Re: Combinations. Confused on the terms used in a solution.
How are you getting 560 out of that? And do they have 6 choices for storage or 5?

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• Jun 19th 2012, 07:54 AM
polcareb192
Re: Combinations. Confused on the terms used in a solution.
5 heights to choose from for 4 modules. 5C1 note[all 4 modules are same height] 2 x 5C2 note[3 modules of the same height, 1 module a different height (of the 5 heights choose 2 different heights, of the one chosen height you must use 3 of them for the storage module height of the first 3 storage modules and from the other height chosen you get the 4th height for the 4th storage module) Multiply by 2 because It can be AAAB or BAAA to get 1 the same and 3 the same, so it is 2 X 5C2]
• Jun 19th 2012, 08:08 AM
polcareb192
Re: Combinations. Confused on the terms used in a solution.
5C2 note[two different sized pairs, so choose 2 different sizes AABB for example (so you chose A and B)]
5C3 X 3 note[a pair and 2 single heights (AABC, BBAC, CCAB with each of 3 sizes being the pair or not so multiply 5C3 by those 3 sizes)
5C4 note[of the 5 heights you choose 4, all different heights, so ABCD for example]
So you have, finally:
(5C1) 2(5C2) (5C2) 3(5C3) (5C4) (4) (2) note the 4 is the amount of tops to choose from and 2 is the amount of bases to choose from
multiply above terms to get 560.