Showing posts with label combinations. Show all posts
Showing posts with label combinations. Show all posts

Saturday, August 4, 2012

Exploring Exponential Expansion: Combinations!

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I am fascinated, as are most people, at rates of expansion, growth and change. Albert Einstein once said (in a Swiss accent whilst tripping over an untied shoelace) that "the most powerful force in the universe is the compounding of interest."

The idea of arithmetic, geometric, exponential and logarithmic expansion, looked at singly or comparatively, give us accessible, understandable metrics about how much things change as a result of the rate or power of growth. Everyone has an interest in these things...from the little tike saving up his meager allowance for his first WonkaBar lottery ticket purchase, to the epidemiologists (mostly the ones in the shows on late night television anyway) who must estimate the potential for a virus or other threatening bio hazard to spread, the the idiot from some fourth-world nation who sends out a chain letter to a whole group of people saying to each of them that if he or she sends the letter to an additional seven people enclosing one [insert any imaginary unit of currency], he or she will become a billionaire [i.e., that he or she will receive one billion units of the imaginary units of currency] within seven weeks - or, alternatively, that if he or she (poor recipient) fails to  do as instructed (gasp!!!), that he or she will be visited by the angel of halitosis and be breathed upon until his or her skin is covered with festering boils...or worse.

One area worth exploring for all those of you involved in business and networking, or search engine optimization and link-building, is that of combinatorial analysis. This is the process through which we answer the question "How many combinations of r objects each are possible out of a total group of n objects.

The formula used to solve this type of problem is:


where n is the number of things to choose from, and you choose r of them
(No repetition, order matters)

As a practical, yet tepid and unimaginative example, let us assume that We are invited to a business power breakfast at the local Olympia Diner. If there are ten invitees in total, and each person shakes hands once with every other person at the end of the meeting (without anyone  leaving a tip for Rayette, the waitress with a prominent mole who calls everybody "hon' "), how many handshakes will there be in all?

This question, restated without the cynical editorial commentary, is simply, "How many combinations of 2 are possible out of a group of 10, in total?"

Using the above formula, we take 10!, and divide it by (10! - 2!), simplify it to 10! divided by 8!, further simplify to 10 x 9, and we obtain an answer of 90 possible handshakes.

By the way, if you need an automatic full factorial calculator just click *HERE*. I never leave the house without mine. It beats trying to figure out 123! either in my head (which is rather full) or on the back of all of my wife's best linen table napkins with my Bic.

Here's one for You:

If I own fourteen suits, and I wear two different suits during any given week, how many different combinations of suits will I be able to wear?

Here's another for You, with a slight twist:

If I publish 50 different blogs, and I reference all of the other blogs on each one of the blogs, how many total links will I have to any one given blog on each of the others, (inclusive of that blog itself)? Hmmm? And how many links will I have which link one blog to another?

Stay tuned for the answers.

Douglas E. Castle for The Braintenance Blog






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Monday, April 5, 2010

Permutations and Combinations - The Answers!

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Dear Friends:

Here are the answers (long-awaited) for the questions posed las week which involve permutations and combinations:

1. In a group of ten individuals, how many pairings are possible?


Based upon the formula (you might want to roll back a few posts to find it), the answer is 10!/(10! - 2!)(2!), which equals 10 x 9 /2 which equals 45 possible pairings.

2. In a group of ten individuals, how many combinations of 3 individuals each are possible?

Using the same formula as above, the answer is 10 x 9 x 8 / 3, which equals 168 possible combinations of 3 each.

3. In a group of ten individuals, how many combinations of 2 or 3 individuals are possible?

The key word is "or." Merely add the answer to question 1 to the answer to question 2, or simply add 45 + 168, which equals 213 possible combinations of either 2 or 3 each.

4. Five paintings are to be arranged in a row on a wall. How many different arrangements are possible?

Using the permutation formula, the answer can be found by multiplying 5 x 4 x 3 x 2 x 1, which will equal 120 possible arrangements.



5. Same facts as in question 4, above: How many different arrangements are possible where two paintings by one particular artist must be side by side?

If one artist produced two fo the paintings, and he or she would like both of those paintings displayed side by side, we will have less than 120 possible arrangements. There are several arrangements possible where two paintings MUST be side by side:

1 x 1 x 3 x 2 x 1 =  6 arrangements

3 x 1 x 1 x 2 x 1 =  6 arrangements

3 x 2 x 1 x 1 x 1 =  6 arrangements

3 x 2 x 1 x 1 x 1 =  6 arrangements

In all, there are 6 + 6 + 6 + 6 total arrangements, or a total of 34 possible arrangements where the demanding artist will have his or her two works hanging side by side.


Faithfully,

Douglas Castle
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Monday, March 29, 2010

Braintenance: Permutations and Combinations!

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Dear Fellow Cranial Crunchers:

You now have some problems. Wait. Let me re-phrase that. It was politically incorrect. Perhaps, "You now have some issues." No... perhaps, "You must now address some challenges." Ah, much better.

1. In a group of ten individuals, how many pairings are possible?

2. In a group of ten individuals, how many combinations of 3 individuals each are possible?

3. In a group of ten individuals, how many combinations of 2 or 3 individuals are possible?

4. Five paintings are to be arranged in a row on a wall. How many different arrangements are possible?

5. Same facts as in question 4, above: How many different arrangements are possible where two paintings by one particular artist must be side by side?

Hop to it!

Faithfully,

Douglas Castle
--------------------------
For some helpful hints to assist you in finding the right answers for the above questions, you can get a free instructive download at:
http://www.mediafire.com/?ldfhznjmgmy


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Comments to Douglas Castle
About Douglas Castle 
Douglas Castle - LinkedIn Profile
The National Networker Companies
Braintenance - Stay razor sharp.
The Internationalist Page - A world without barriers.
The Global Futurist - Revealing trends.
Taking Command! - Mastering your fate.
Follow Castle on Twitter 
Follow TNNW on Twitter
Follow Braintenance on Twitter
*Subscribe (free!) for The National Networker Newsletter and the BLUE TUESDAY REPORT, and join The TNNWC GICBC at Join Us!

Friday, January 23, 2009

BRAINTENANCE QUIZ 1/23/09

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Dear Friends:


Now that you have had the opportunity to enjoy a well-earned day's rest, let's switch the format of our meeting. I offer some solutions before I pose new problems. Variety is an intellectual stimulant (as it is a leading cause for divorce in the US...but then, that's in a different context). Please refer to the last two posts on this blog, as it's generally a good policy to know the questions tha the answers relate to!
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Question: What would the amount and the dilution be of hyroxyidiotic acid be in bottle #3 at the end of this process? This is a question that can be answered intuitively, without much math at all (phew). In Question #3, we were left with 1 gallon of 20% hydroxyidiotic acid in bottle #3. In Question #4, we added 1 gallon of 10% of hydroxyidiotic acid to bottle #3. In sum, bottle #3 now contains 1 gallon of 20% acid, and 1 gallon of 10% acid, or 2 gallons of 15% hydroxyidiotic acid. If we were to have done the math, we might have used this calculation: (.50) (.20) +(.50) (.10) = .15 , or 15%.
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How many 5-letter combinations can you make of the name OBAMA, where none of those combinations started with an "O"? The computation is simple (but with a twist): for the first letter slot you have 4 possible letters; for the second, 4; for the third, 3: for the fourth, 2; and for the fifth, 1 --- by multiplying 4x4x3x2x1, we get 96 arrangements of letters that do not begin with the letter "O". But what about the arrangements that are redundant because they have two "A"s in them....Hmmm....perhaps you need a bit more time.

Since every arrangement of these letters will contain two letter "A"s, and it doesn't matter which A is being used, half of our arrangements will be redundant. What we must then do is eliminate half of the arrangements in order to eliminate these redundancies. If we divide 96 in half, the result will be 48 arrangements, which is our answer.
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Now for a second surprise...there is no problem for today. There is merely a wonderful exercise. Take at least 30 minutes at some time time today, and sample some of the audio/video clips below for mind expansion purposes. Scroll down to EXERCISE YOUR MIND 6: MEDITATION, STIMULATION AND RELAXATION. Enjoy the experience!

Faithfully,

Douglas Castle
INTERNAL ENERGY PLUS

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