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 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:
 ```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 http://mathforum.org/dr.math/faq/faq.comb.perm.html http://www.themathpage.com/aPreCalc/permutations-combinations.htm 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: http://68.15.21.151/uploads/researchers/magazine.txt 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. Thanks, smudgy.``` 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. Thanks, smudgy.```
 singingleaves-ga rated this answer: `Exactly what I needed and a quick response. Thank you.`

 ```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!```
 ```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 magazines. 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. http://mathworld.wolfram.com/PascalsTriangle.html```