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Q: Mathamatical Permutations for Business ( Answered 5 out of 5 stars,   2 Comments )
Subject: Mathamatical Permutations for Business
Category: Science > Math
Asked by: singingleaves-ga
List Price: $50.00
Posted: 10 Aug 2004 18:56 PDT
Expires: 09 Sep 2004 18:56 PDT
Question ID: 386192
I have six magazines. I plan to add a seventh in two months.
Advertisers may purchase an ad in just one publication or any
combination of the seven, including all seven. Q. How many
combinations of 2, 3, 4, 5, & 6 publications are there WITHOUT
duplicating combinations? And what are these publication combinations?
Publications may be referred to as MT, YV, NG, NH, LL, SS, WC
I need a listing of each possible option... I do not need MT & YV...YV
& MT which are the same combination.
Subject: Re: Mathamatical Permutations for Business
Answered By: smudgy-ga on 10 Aug 2004 21:24 PDT
Rated:5 out of 5 stars
Hi singingleaves,

I hope you find the following answer satisfactory. If not, please
request a clarification before rating and I will do my best to resolve
any issues.

There are 119 ways to choose two, three, four, five, or six magazines
from seven in such a way that order is unimportant.

In order to perform this calculation, we use the mathematical idea of
combinations. The question you ask is basically akin to asking, "If
there are seven available pizza toppings, how many pizzas can we order
with anywhere from two to six toppings?"

To give a quick example of a specific one of the cases we're looking
at: let's say we are putting three out of seven toppings on our pizza.
(This is denoted "7 choose 3" or "7C3" mathematically.) Then there are
seven choices for the first topping, six choices for the second, and
five choices for the third. This seems to indicate that there are
7*6*5=210 possible pizzas we can make. BUT, it doesn't matter what
order we put the toppings on--if we put on onions, then mushrooms,
then peppers, this is the same pizza that we have if we put on
mushrooms, then peppers, then onions. There are -six- different orders
that we could have applied the three chosen toppings, all of which
make the same pizza in the end, so our original guess of 210 different
pizzas is too great by a factor of six. So the actual number of
three-topping pizzas is 210/6=35.

For more information on mathematical combinations and how to calculate them, see

Here are the number of available combinations of 2, 3, 4, 5, and 6
magazines from seven:

Choose two magazines from seven: 21 ways
Three from seven: 35 ways
Four from seven: 35 ways
Five from seven: 21 ways
Six from seven: 7 ways

For a total of 119 ways to choose 2, 3, 4, 5, or 6 magazines from seven.

Additionally, there are seven ways to choose just one magazine, one
way to choose all seven magazines, and one way to choose no magazines
at all.

I have written out a list of all the possible combinations, but it is
quite long, so I will point you to a text file:

I hope this answers your question! If you have any further questions,
or if you cannot access the text file, please request a clarification
before rating and I will do my best to clarify my answer.


Clarification of Answer by smudgy-ga on 10 Aug 2004 21:41 PDT
I should add:

Google search terms:
permutation combination

This search will pull plenty of information on calculating
permutations and combinations, from a variety of sources.

singingleaves-ga rated this answer:5 out of 5 stars
Exactly what I needed and a quick response. Thank you.

Subject: Re: Mathamatical Permutations for Business
From: nemtudom-ga on 10 Aug 2004 21:47 PDT
Wow! Singingleaves, you're paying $50 for the answer to *one* homework
problem? At that rate, you must be spending as much money to keep from
learning anything as you're spending on tuition!
Subject: A different answer?
From: ulu-ga on 11 Aug 2004 02:38 PDT
Rereading your question and the preamble, the "simpler" question is
how many different ways of picking a nonzero set from the set of all 7

You can think of all the combinations as:  does the magazine have the
ad or not (a binary question).

That would be 2*2*2*2*2*2*2 - the cases where none or all have an ad.

You end up with 2^7 - (1+1) = 128-2 = 126.

Your official question also removed the case of just one ad:  126 - 7 = 119

All rows of Pascal's triangle add up to a power of 2.

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